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All questions of Gravitation for Class 11 Exam
Two bodies with same mass “m” separated by a distance “r” exert a gravitational force of F on each other. Suppose the distance between them is doubled and the force becomes F’. The ratio of two forces is- a)1:4
- b)4:1
- c)1:2
- d)2:1
Correct answer is 'B'. Can you explain this answer?
Two bodies with same mass “m” separated by a distance “r” exert a gravitational force of F on each other. Suppose the distance between them is doubled and the force becomes F’. The ratio of two forces is
a)
1:4
b)
4:1
c)
1:2
d)
2:1
|
Niharika Nair answered |
We know that the force of gravitation is inversely proportional to square of the distance between the two bodies,
i.e. F∝ r-2
Hence, when the distance between them will be doubled, the force will be reduced by 4 times
So, the ratio will be 4:1
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of V.- a)Ve = 2V
- b)Ve= V
- c)Ve= V/2
- d)Ve =
Correct answer is option 'A'. Can you explain this answer?
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of V.
a)
Ve = 2V
b)
Ve= V
c)
Ve= V/2
d)
Ve =
|
Riya Banerjee answered |
What does the term escape speed of a planet mean?- a)the minimum speed of an object to reach a height equal to the radius of the planet
- b)cannot be defined
- c)the minimum speed required for an object to reach infinity
- d)the maximum speed required for an object to reach infinity
Correct answer is option 'C'. Can you explain this answer?
What does the term escape speed of a planet mean?
a)
the minimum speed of an object to reach a height equal to the radius of the planet
b)
cannot be defined
c)
the minimum speed required for an object to reach infinity
d)
the maximum speed required for an object to reach infinity
|
Om Desai answered |
Escape velocity is the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.
Can you explain the answer of this question below:The value of escape velocity on earth when the body is thrown making an angle of 45o with the horizontal is
- A:
11.2 x 21/4 km/s
- B:
22.4 km/s
- C:
11.2 x 21/2 km/s
- D:
11.2 km/s
The answer is d.
The value of escape velocity on earth when the body is thrown making an angle of 45o with the horizontal is
11.2 x 21/4 km/s
22.4 km/s
11.2 x 21/2 km/s
11.2 km/s
|
Neha Joshi answered |
Escape velocity does not depend on the angle of projection. Escape velocity will remain the same.
Hence, escape velocity is 11.2km/s .
Hence, escape velocity is 11.2km/s .
Which is untrue about orbital velocity?
- a)increases with the increase in height of satellite
- b)depends on mass and radius of planet around which it revolves
- c)it is independent of mass of satellite
- d)decreases with an increase in radius of orbit
Correct answer is option 'A'. Can you explain this answer?
Which is untrue about orbital velocity?
a)
increases with the increase in height of satellite
b)
depends on mass and radius of planet around which it revolves
c)
it is independent of mass of satellite
d)
decreases with an increase in radius of orbit
|
|
Om Desai answered |
The untrue statement about orbital velocity is:
1. increases with the increase in height of satellite
Explanation: Orbital velocity is the speed at which an object revolves around a planet or other celestial body in a stable orbit. According to the equation for orbital velocity, v = √(GM/r+h), where G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit.
As the height of the satellite increases (meaning it gets far to the planet), its h increase , so reasulting in decrease in velocity .
The other statements are true:
2. depends on mass and radius of planet around which it revolves: As mentioned in the equation, orbital velocity depends on both the mass (M) of the planet and the radius (r) of the orbit.
3. it is independent of mass of satellite: The mass of the satellite does not appear in the equation for orbital velocity, so it does not affect the speed at which the satellite orbits the planet.
4. decreases with an increase in radius of orbit: From the equation, we can see that as the radius of the orbit (r) increases, the orbital velocity (v) decreases.
1. increases with the increase in height of satellite
Explanation: Orbital velocity is the speed at which an object revolves around a planet or other celestial body in a stable orbit. According to the equation for orbital velocity, v = √(GM/r+h), where G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit.
As the height of the satellite increases (meaning it gets far to the planet), its h increase , so reasulting in decrease in velocity .
The other statements are true:
2. depends on mass and radius of planet around which it revolves: As mentioned in the equation, orbital velocity depends on both the mass (M) of the planet and the radius (r) of the orbit.
3. it is independent of mass of satellite: The mass of the satellite does not appear in the equation for orbital velocity, so it does not affect the speed at which the satellite orbits the planet.
4. decreases with an increase in radius of orbit: From the equation, we can see that as the radius of the orbit (r) increases, the orbital velocity (v) decreases.
Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.Kinetic energy of the system just after the collision is- a)
- b)
- c)
- d)Cannot be determined
Correct answer is option 'B'. Can you explain this answer?
Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.
Kinetic energy of the system just after the collision is
a)
b)
c)
d)
Cannot be determined
|
Lavanya Menon answered |
India’s first artificial satellite was:- a)Sputnik
- b)Rohini
- c)Insat
- d)Aryabhatta
Correct answer is option 'D'. Can you explain this answer?
India’s first artificial satellite was:
a)
Sputnik
b)
Rohini
c)
Insat
d)
Aryabhatta
|
Geetika Shah answered |
India's first artificial satellite was Aryabhata. Launched by soviet union on 19 April 1975
A geostationary satellite is at a height h above the surface of earth. If earth radius is R 
- a)The minimum colatitude on earth upto which the satellite can be used for communication is sin_1 (R/R + h)
- b)The maximum colatitudes on earth upto which the satellite can be used for communication is sin_1 (R/R + h)
- c)The area on earth escaped from this satellite is given as 2pR2(1 + sinq)
- d) The area on earth escaped from this satellite is given as 2pR2(1 + cosq)
Correct answer is option 'A,C'. Can you explain this answer?
A geostationary satellite is at a height h above the surface of earth. If earth radius is R
a)
The minimum colatitude on earth upto which the satellite can be used for communication is sin_1 (R/R + h)
b)
The maximum colatitudes on earth upto which the satellite can be used for communication is sin_1 (R/R + h)
c)
The area on earth escaped from this satellite is given as 2pR2(1 + sinq)
d)
The area on earth escaped from this satellite is given as 2pR2(1 + cosq)
|
Krishna Iyer answered |
Minimum colatitude=sin θ=R/R+h
θ=sin-1(R/R+h)
The curved area AB on earth =2πR2(1−sinθ)
Area on earth escaped from satellite =4πR2−2πR2(1−sinθ)
=2πR2(1+sinθ)
θ=sin-1(R/R+h)
The curved area AB on earth =2πR2(1−sinθ)
Area on earth escaped from satellite =4πR2−2πR2(1−sinθ)
=2πR2(1+sinθ)
An unmanned space probe is to the thrown with such a velocity that it does not return to earth. It should be thrown with- a)velocity ≥ (2GM/ R)1/2
- b)a great velocity but nothing more can be said
- c)velocity >(2GM)1/2/ R necessarily
- d)only the velocity = (2GM)1/2/R
Correct answer is option 'A'. Can you explain this answer?
An unmanned space probe is to the thrown with such a velocity that it does not return to earth. It should be thrown with
a)
velocity ≥ (2GM/ R)1/2
b)
a great velocity but nothing more can be said
c)
velocity >(2GM)1/2/ R necessarily
d)
only the velocity = (2GM)1/2/R
|
|
Riya Banerjee answered |
It should be thrown with a velocity greater than or equal to escape velocity of the planet only then it will not return to Earth
Escape Velocity = √(2GM/R) = (2GM/R)1/2
Hence A is the correct answer.
Escape Velocity = √(2GM/R) = (2GM/R)1/2
Hence A is the correct answer.
Assuming the earth to be a sphere of uniform density the acceleration due to gravity- a)At a point outside the earth is inversely proportional to the square of its distance from the center
- b) At a point outside the earth is inversely proportional to its distance from the centre
- c)At a point inside is zero
- d)At a point inside is proportional to its distance from the centre
Correct answer is option 'A,D'. Can you explain this answer?
Assuming the earth to be a sphere of uniform density the acceleration due to gravity
a)
At a point outside the earth is inversely proportional to the square of its distance from the center
b)
At a point outside the earth is inversely proportional to its distance from the centre
c)
At a point inside is zero
d)
At a point inside is proportional to its distance from the centre
|
|
Krishna Iyer answered |
As g is inversely proportional to (R+h)2 when we go away from earth's surface g is inversely proportional to square of the distance and g is directly proportional to (R-d) when we go inside the surface of Earth therefore g is directly proportional to distance travelled inside the Earth
g=Gmr/a3=r for r<R
g=Gm/a2 for r<R
g=Gmr/a3=r for r<R
g=Gm/a2 for r<R
The escape velocity for the moon is nearly- a)11.2km/s
- b)2.4km/s
- c)24km/s
- d)10km/s
Correct answer is option 'B'. Can you explain this answer?
The escape velocity for the moon is nearly
a)
11.2km/s
b)
2.4km/s
c)
24km/s
d)
10km/s
|
Rohit Shah answered |
About 11.2 km/s
In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s).
A satellite which appears to be at a fixed position at a definite height to an observer is called:- a)Geostationary satellite and geosynchronous satellite
- b)Polar satellite
- c)Geostationary satellite
- d)Geosynchronous satellite
Correct answer is option 'A'. Can you explain this answer?
A satellite which appears to be at a fixed position at a definite height to an observer is called:
a)
Geostationary satellite and geosynchronous satellite
b)
Polar satellite
c)
Geostationary satellite
d)
Geosynchronous satellite
|
|
Neha Joshi answered |
As the relative velocity of the satellite with respect to the earth is zero, it appears stationary from the Earth surface and therefore it is called is geostationary satellite or geosynchronous satellite.
A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is- a)Mg/h
- b)
mgh - c)
mgh - d)
mgh
Correct answer is option 'B'. Can you explain this answer?
A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is
a)
Mg/h
b)
mgh c)
mghd)
mgh
|
Suresh Iyer answered |
Initial PE = -GMm/R
Final PE = -GMm/R+h, (h=R/5)
Increase in PE = final - initial
= -GMm/R + (R/5) - (-GMm/R)
=GMm(1/R - 5/6R)
=GMm/R * 1/6
= gmR/6 (g= GM/R)
=gm5h/6 (h=R/5)
Final PE = -GMm/R+h, (h=R/5)
Increase in PE = final - initial
= -GMm/R + (R/5) - (-GMm/R)
=GMm(1/R - 5/6R)
=GMm/R * 1/6
= gmR/6 (g= GM/R)
=gm5h/6 (h=R/5)
Weightlessness is:- a)a situation in which the effective weight of the body becomes zero
- b)a situation in which the effective weight of the body becomes 10N
- c)a situation in which the effective weight of the body becomes 9.8N.
- d)a situation in which the effective mass of the body becomes zero
Correct answer is option 'A'. Can you explain this answer?
Weightlessness is:
a)
a situation in which the effective weight of the body becomes zero
b)
a situation in which the effective weight of the body becomes 10N
c)
a situation in which the effective weight of the body becomes 9.8N.
d)
a situation in which the effective mass of the body becomes zero
|
Akshay Shah answered |
Weight becomes zero in a freefall or when the total force on your body is zero.For example in a freely falling lift the contact force on your body is zero,so the wieghting machine would show zero reading.so you are wieghtless.
The gravitational potential due to the gravitational force on the earth is defined as the- a)potential energy multiplied by the mass of the object
- b)potential energy of the mass placed at that point
- c)numerically equal to the potential energy
- d)potential energy of a unit mass at that point.
Correct answer is option 'D'. Can you explain this answer?
The gravitational potential due to the gravitational force on the earth is defined as the
a)
potential energy multiplied by the mass of the object
b)
potential energy of the mass placed at that point
c)
numerically equal to the potential energy
d)
potential energy of a unit mass at that point.
|
Nandini Patel answered |
Gravitational Potential
Gravitational Potential is dened as the potential energy of a particle of unit mass at that point due to the gravitational force exerted byearth. Gravitational potential energy of a unit mass is known as gravitational potential.
What could be the maximum value for gravitational potential energy?- a)1
- b)zero
- c)infinity
- d)1000
Correct answer is option 'B'. Can you explain this answer?
What could be the maximum value for gravitational potential energy?
a)
1
b)
zero
c)
infinity
d)
1000
|
|
Riya Banerjee answered |
As, U=−Gm1m2/r,
it can be used to describe the potential energy in a system of point charges (or radially symmetric spheres) as a function of their separation distance r, then the maximum value is zero at infinite separation.
Hence, its maximum value is zero at infinity.
it can be used to describe the potential energy in a system of point charges (or radially symmetric spheres) as a function of their separation distance r, then the maximum value is zero at infinite separation.
Hence, its maximum value is zero at infinity.
When a satellite moves in a circular orbit, the _______acceleration is provided by the gravitational attraction of the earth- a)tangential
- b)centrifugal
- c)centripetal
- d)fictitious
Correct answer is option 'C'. Can you explain this answer?
When a satellite moves in a circular orbit, the _______acceleration is provided by the gravitational attraction of the earth
a)
tangential
b)
centrifugal
c)
centripetal
d)
fictitious
|
Hansa Sharma answered |
When any body or particle moves in circular orbit centripetal force acts on it and to move the object centripetal acceleration is necessary. So, when a satellite moves in circular orbit the centripetal acceleration is provided by the gravitational attraction of the earth.
Kepler’s second law states that the straight line joining the planet to the sun sweeps out equal areas in equal time. The statement is equivalent to saying that:- a)longitudnal acceleration is zero
- b)total acceleration is zero
- c)transverse acceleration is zero
- d)radial acceleration is zero
Correct answer is option 'C'. Can you explain this answer?
Kepler’s second law states that the straight line joining the planet to the sun sweeps out equal areas in equal time. The statement is equivalent to saying that:
a)
longitudnal acceleration is zero
b)
total acceleration is zero
c)
transverse acceleration is zero
d)
radial acceleration is zero
|
|
Suresh Reddy answered |
According to the second law the orbital radius and angular velocity of the planet in the elliptical orbit will vary. The planet travels faster when closer to the Sun, then slower when farther from the Sun. Hence we can say that the transverse acceleration is zero while radial and longitudinal accelerations are not zero.
Velocity of geostationary satellite with respect to earth is- a)10 ms-1
- b)15 ms-1
- c)zero
- d)1 ms-1
Correct answer is option 'C'. Can you explain this answer?
Velocity of geostationary satellite with respect to earth is
a)
10 ms-1
b)
15 ms-1
c)
zero
d)
1 ms-1
|
|
Neha Joshi answered |
Because geostationary satellite ( completes it's rotation in about 24hrs).and it's rotation cycle is same as that of Earth .thus it's velocity is zero with respect to earth
A ball is thrown vertically upwards. The acceleration due to gravity:- a)is in the upward direction
- b)is in the downward direction
- c)is in the horizontal direction
- d)always in the direction of motion
Correct answer is option 'B'. Can you explain this answer?
A ball is thrown vertically upwards. The acceleration due to gravity:
a)
is in the upward direction
b)
is in the downward direction
c)
is in the horizontal direction
d)
always in the direction of motion
|
|
Gaurav Kumar answered |
A ball is thrown vertically upwards. The acceleration due to gravity is in the downward direction.
The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth ?- a)
second - b)
seconds - c)
second - d)
second
Correct answer is option 'B'. Can you explain this answer?
The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth ?
a)
secondb)
secondsc)
secondd)
second
|
Pooja Shah answered |
As Mp=2Me and Dp=2De
or Rp=2Re
Hence, gp= GMp/(Rp)2 =G(2Me)/(2Re)2 = GMe/2Re2 =ge/2
Time period of pendulum on the planet Tp=2π√ l/gp
Tp=2π√2l/ge=√2×2π√l/ge=√2×Te
Tp=√2×2=2√2s
or Rp=2Re
Hence, gp= GMp/(Rp)2 =G(2Me)/(2Re)2 = GMe/2Re2 =ge/2
Time period of pendulum on the planet Tp=2π√ l/gp
Tp=2π√2l/ge=√2×2π√l/ge=√2×Te
Tp=√2×2=2√2s
An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. which one of following quantity increases- a)gravitational potential energy
- b)centripetal acceleration
- c)gravitationl force
- d)linear orbital speed
Correct answer is option 'A'. Can you explain this answer?
An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. which one of following quantity increases
a)
gravitational potential energy
b)
centripetal acceleration
c)
gravitationl force
d)
linear orbital speed
|
|
Neha Joshi answered |
We know that gravitational potential is negative in sign and its magnitude decreases when distance from the massive attracting object increases, hence when considered with sign we can say that gravitational potential increases with increases in distance.
If a satellite orbits as close to the earth's surface as possible,- a) Its speed is maximum
- b)Time period of its rotation is minimum
- c)The total energy of the `earth plus satellite' system is minimum
- d)The total energy of the `earth plus satellite' system is maximum
Correct answer is option 'A,B,C'. Can you explain this answer?
If a satellite orbits as close to the earth's surface as possible,
a)
Its speed is maximum
b)
Time period of its rotation is minimum
c)
The total energy of the `earth plus satellite' system is minimum
d)
The total energy of the `earth plus satellite' system is maximum
|
|
Hansa Sharma answered |
The velocity of satellite V=√GM/R⇒V∝1/√R.
The time period of satellite is given by T=2π√ R3/GM⇒T∝R1.5
The total energy is given by T.E.=−GMm/2R⇒T.E.∝−1/M
Here, G is constant, R is orbital radius, M is mass of earth and m is mass of satellite.
Near the earth surface, R has the least value and therefore time period and total energy are minimum while velocity is maximum. So, options A, B, C are correct.
The time period of satellite is given by T=2π√ R3/GM⇒T∝R1.5
The total energy is given by T.E.=−GMm/2R⇒T.E.∝−1/M
Here, G is constant, R is orbital radius, M is mass of earth and m is mass of satellite.
Near the earth surface, R has the least value and therefore time period and total energy are minimum while velocity is maximum. So, options A, B, C are correct.
Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is- a)1 : 2
- b)1 : 4
- c)1 : 8
- d)1 : 16
Correct answer is option 'A'. Can you explain this answer?
Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is
a)
1 : 2
b)
1 : 4
c)
1 : 8
d)
1 : 16
|
|
Gaurav Kumar answered |
Answer :- a
Solution :- L = mvr
⇒L=m√GMr/r
L = m√GMr−−−−.....(1)
L = 2mdA/dt....(2)
From (1) and (2)
dA/dt ∝ (r)^1/2
= √(dA/dt)1/(dA/dt)2
= √4/1 = 2/1
Gravitational Potential is:- a)negative, scalar quantity , unit JKg-1
- b)positive, vector quantity , unit JKg-1
- c)positive, scalar quantity , unit JKg-1
- d)negative, vector quantity , unit JKg-1
Correct answer is option 'A'. Can you explain this answer?
Gravitational Potential is:
a)
negative, scalar quantity , unit JKg-1
b)
positive, vector quantity , unit JKg-1
c)
positive, scalar quantity , unit JKg-1
d)
negative, vector quantity , unit JKg-1
|
Ayush Joshi answered |
Gravitational potential (radial fields) at a point in a radial field is the work done per unit mass against the field, in bringing a small mass from infinite distance to the point. Since gravitational fields are attractive and the potential at infinite distance is zero, all points within the field have negative values of potential. Gravitational potential is a scalar quantity with SI unit J kg^-1. The symbol used is mostly V but sometimes or Vr or V(r). A radial gravitational field is one in which the field strength has the same magnitude at all points at a given distance from the center.
The energy required to be spent by a satellite of mass m and speed v and orbital radius r in completing a circular orbit once round the earth of mass M is- a)GMm/r
- b)2GMm/r
- c)GMm/2r
- d)zero
Correct answer is 'D'. Can you explain this answer?
The energy required to be spent by a satellite of mass m and speed v and orbital radius r in completing a circular orbit once round the earth of mass M is
a)
GMm/r
b)
2GMm/r
c)
GMm/2r
d)
zero
|
|
Suresh Iyer answered |
As the satellite is bounded by earth's gravitational field and has some velocity, it would completely revolve around the circular orbit without any external energy required.
A hollow spherical shell is compressed to half its radius. The gravitational potential at the centre- a)Increases
- b) Decreases
- c) Remains same
- d)Ruring the compression increases then returns at the previous value
Correct answer is option 'B'. Can you explain this answer?
A hollow spherical shell is compressed to half its radius. The gravitational potential at the centre
a)
Increases
b)
Decreases
c)
Remains same
d)
Ruring the compression increases then returns at the previous value
|
|
Suresh Iyer answered |
Gravitational Potential V = -GM/R for hollow spherical shell at the centre. If we replace R by R/2 then we get V = -2GM/R. Therefore it decreases.
If a tunnel is cut at any orientation through earth, then a ball released from one end will reach the other end in time (neglect earth rotation)- a)84.6 minutes
- b)42.3 minutes
- c)8 minutes
- d)Depends on orientation
Correct answer is option 'B'. Can you explain this answer?
If a tunnel is cut at any orientation through earth, then a ball released from one end will reach the other end in time (neglect earth rotation)
a)
84.6 minutes
b)
42.3 minutes
c)
8 minutes
d)
Depends on orientation
|
|
Neha Joshi answered |
Total time period will be 84.6 min when ball released from one end and it come backs to same point as in oscillation.
When ball is released from one end then time taken to reach other end will be half of total time period then that will be 42.3 min.
When ball is released from one end then time taken to reach other end will be half of total time period then that will be 42.3 min.
When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then- a) Its kinetic energy increases
- b)Its kinetic energy decreases
- c)Its angular momentum about the earth decreases
- d)Its period of revolution around the earth increases
Correct answer is option 'A'. Can you explain this answer?
When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then
a)
Its kinetic energy increases
b)
Its kinetic energy decreases
c)
Its angular momentum about the earth decreases
d)
Its period of revolution around the earth increases
|
|
Hansa Sharma answered |
Due to air drag some mechanical energy of satellite will converted into heat energy, there will be loss of ME of satellite, so radius of orbit will decrease and satellite will follow a spiral path towards earth.
If r is decreases,
P.E.=−GMm/r, P.E. is also decreases.
K.E=GMm/2r K.E increasing, increased speed in spite of decrease in M.E. But rate of P.E decreases is more than the rate at which M.E decreases.
Angular momentum L=mrv=mr√GMr=m√GMr⇒L∝√r.
So, angular momentum decreases as r decreases.
If r is decreases,
P.E.=−GMm/r, P.E. is also decreases.
K.E=GMm/2r K.E increasing, increased speed in spite of decrease in M.E. But rate of P.E decreases is more than the rate at which M.E decreases.
Angular momentum L=mrv=mr√GMr=m√GMr⇒L∝√r.
So, angular momentum decreases as r decreases.
Escape velocity is:- a)The minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.
- b)The minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and returns back
- c)The maximum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and returns back.
- d)The maximum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own
Correct answer is option 'A'. Can you explain this answer?
Escape velocity is:
a)
The minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.
b)
The minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and returns back
c)
The maximum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and returns back.
d)
The maximum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own
|
|
Geetika Shah answered |
Escape velocity is the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.
A communications Earth satellite- a)Goes round the earth from east to west
- b)Can be in the equatorial plane only
- c)Can be vertically above any place on the earth
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A communications Earth satellite
a)
Goes round the earth from east to west
b)
Can be in the equatorial plane only
c)
Can be vertically above any place on the earth
d)
none of these
|
|
Pooja Shah answered |
Communication satellites are supposed to be geostationary in nature. So, from the above properties options B is correct.
The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is- a) R/2
- b)2R
- c)R/3
- d)3R
Correct answer is option 'B'. Can you explain this answer?
The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is
a)
R/2
b)
2R
c)
R/3
d)
3R
|
|
Bhargavi Sen answered |
To find the height at which the acceleration due to gravity becomes g/9, let's analyze the gravitational force equation.
Gravitational force (F) between two objects is given by:
F = (G * m1 * m2) / r^2
where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
The acceleration due to gravity (g) is the force experienced by an object of mass m due to the gravitational attraction of the Earth. It is given by:
g = (G * M) / R^2
where M is the mass of the Earth and R is the radius of the Earth.
Since the acceleration due to gravity is directly proportional to the mass of the Earth and inversely proportional to the square of the radius, we can write:
g' = (G * M') / (R')^2
where g' is the new acceleration due to gravity, M' is the new mass of the Earth, and R' is the new radius of the Earth.
We are given that g' = g/9. Substituting this into the equation above, we get:
g/9 = (G * M') / (R')^2
Simplifying, we find:
(R')^2 = (9 * M' * R^2) / (G * M)
To find the height at which the acceleration due to gravity becomes g/9, we need to find the difference between the new radius (R') and the original radius (R). Let's call this difference ΔR.
ΔR = R' - R
Now, let's substitute the expression for (R')^2 into the equation for ΔR:
ΔR = sqrt((9 * M' * R^2) / (G * M)) - R
Since the new mass M' is the same as the original mass M (since we are still considering the Earth), we can simplify further:
ΔR = sqrt((9 * R^2) / G) - R
To find the height, we need to multiply ΔR by the radius of the Earth, R:
Height = R * ΔR
Substituting the expression for ΔR, we get:
Height = R * (sqrt((9 * R^2) / G) - R)
Simplifying further, we find:
Height = R * (3R / sqrt(G) - R)
Since G is a constant, the height can be written as:
Height = 3R - R^2 / sqrt(G)
Therefore, the height at which the acceleration due to gravity becomes g/9 is given by 3R - R^2 / sqrt(G), which is equivalent to option B, 2R.
Gravitational force (F) between two objects is given by:
F = (G * m1 * m2) / r^2
where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
The acceleration due to gravity (g) is the force experienced by an object of mass m due to the gravitational attraction of the Earth. It is given by:
g = (G * M) / R^2
where M is the mass of the Earth and R is the radius of the Earth.
Since the acceleration due to gravity is directly proportional to the mass of the Earth and inversely proportional to the square of the radius, we can write:
g' = (G * M') / (R')^2
where g' is the new acceleration due to gravity, M' is the new mass of the Earth, and R' is the new radius of the Earth.
We are given that g' = g/9. Substituting this into the equation above, we get:
g/9 = (G * M') / (R')^2
Simplifying, we find:
(R')^2 = (9 * M' * R^2) / (G * M)
To find the height at which the acceleration due to gravity becomes g/9, we need to find the difference between the new radius (R') and the original radius (R). Let's call this difference ΔR.
ΔR = R' - R
Now, let's substitute the expression for (R')^2 into the equation for ΔR:
ΔR = sqrt((9 * M' * R^2) / (G * M)) - R
Since the new mass M' is the same as the original mass M (since we are still considering the Earth), we can simplify further:
ΔR = sqrt((9 * R^2) / G) - R
To find the height, we need to multiply ΔR by the radius of the Earth, R:
Height = R * ΔR
Substituting the expression for ΔR, we get:
Height = R * (sqrt((9 * R^2) / G) - R)
Simplifying further, we find:
Height = R * (3R / sqrt(G) - R)
Since G is a constant, the height can be written as:
Height = 3R - R^2 / sqrt(G)
Therefore, the height at which the acceleration due to gravity becomes g/9 is given by 3R - R^2 / sqrt(G), which is equivalent to option B, 2R.
A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in- a)Bound orbit
- b)Unbound orbit
- c)Partially bound orbit
- d)Data is insufficient to determine the nature of the orbit
Correct answer is option 'B'. Can you explain this answer?
A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in
a)
Bound orbit
b)
Unbound orbit
c)
Partially bound orbit
d)
Data is insufficient to determine the nature of the orbit
|
|
Suresh Reddy answered |
Applying the conservation of momentum, we get:
V = velocity of 5M, V1 = velocity of 4M, V2 = velocity of M and ATQ V2 = -V
5MV = 4MV1 + M(-V)
6MV = 4MV1
V1 = 3/2 V
Now, Vo = orbital velocity and Ve = escape velocity
Ve = √2 Vo
In a bound orbit the object is gravitationally bound to the body that is the source of gravity (like the Earth is bound to the Sun, or the Moon to the Earth). An unbound orbit is typically hyperbolic, and the object will escape from the source of gravity
V = velocity of 5M, V1 = velocity of 4M, V2 = velocity of M and ATQ V2 = -V
5MV = 4MV1 + M(-V)
6MV = 4MV1
V1 = 3/2 V
Now, Vo = orbital velocity and Ve = escape velocity
Ve = √2 Vo
In a bound orbit the object is gravitationally bound to the body that is the source of gravity (like the Earth is bound to the Sun, or the Moon to the Earth). An unbound orbit is typically hyperbolic, and the object will escape from the source of gravity
The squares of the periods of revolution of the planets are proportional to the ___________of their semimajor axis of its orbit.- a)sqaures
- b)sqaures root
- c)cube
- d)cube roots
Correct answer is option 'C'. Can you explain this answer?
The squares of the periods of revolution of the planets are proportional to the ___________of their semimajor axis of its orbit.
a)
sqaures
b)
sqaures root
c)
cube
d)
cube roots
|
|
Lavanya Menon answered |
The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth- a)The acceleration of S is always directed towards the centre of the earth
- b)The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
- c)The total mechanical energy of S varies periodically with time
- d)The linear momentum of S remains constant in magnitude
Correct answer is option 'A'. Can you explain this answer?
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth
a)
The acceleration of S is always directed towards the centre of the earth
b)
The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
c)
The total mechanical energy of S varies periodically with time
d)
The linear momentum of S remains constant in magnitude
|
|
Bhavana Tiwari answered |
As the gravitational force on satellites due to earth acts always towards the centre of earth, thus acceleration of S is always directed towards the centre of the earth.
Also, as there is no external force so according to conservation of energy, total mechanical energy of S is constant always.
Also, as in the absence of external torque L is constant in magnitude and direction.
Thus, mrv=constant ⟹v varies as r changes
Hence, p=mv is not constant.
Also, as there is no external force so according to conservation of energy, total mechanical energy of S is constant always.
Also, as in the absence of external torque L is constant in magnitude and direction.
Thus, mrv=constant ⟹v varies as r changes
Hence, p=mv is not constant.
The value of acceleration due to gravity on earth is.- a)9.8ms−2.
- b)15376
- c)227004
- d)127008
Correct answer is option 'A'. Can you explain this answer?
The value of acceleration due to gravity on earth is.
a)
9.8ms−2.
b)
15376
c)
227004
d)
127008
|
|
Gaurav Kumar answered |
The value of acceleration due to gravity on earth is g=9.8ms−2.
Hence,
option A is correct answer.
The earth's gravitational force at someplace in space causes an acceleration of 7m/s2 in a 1kg mass.What will be the acceleration of a 5kg mass at the same place?- a)7m/s2
- b)35m/s2
- c)1.4m/s2
- d)3.5m/s2
Correct answer is option 'A'. Can you explain this answer?
The earth's gravitational force at someplace in space causes an acceleration of 7m/s2 in a 1kg mass.What will be the acceleration of a 5kg mass at the same place?
a)
7m/s2
b)
35m/s2
c)
1.4m/s2
d)
3.5m/s2
|
|
Hansa Sharma answered |
The acceleration due to gravity does not vary with the change in the mass of the object as it is a dimensionless quantity representing the amount of matter in a particle or object.
Therefore, the acceleration of a 5 kg mass at the same place is 7m/s2.
Therefore, the acceleration of a 5 kg mass at the same place is 7m/s2.
Two balls are dropped from the same height from places A and B. The body at B takes two seconds less to reach the ground at B strikes the ground with a velocity greater than at A by 10m/s. The product of the acceleration due to gravity at the two places A and B is:- a)5
- b)25
- c)125
- d)12.5
Correct answer is option 'B'. Can you explain this answer?
Two balls are dropped from the same height from places A and B. The body at B takes two seconds less to reach the ground at B strikes the ground with a velocity greater than at A by 10m/s. The product of the acceleration due to gravity at the two places A and B is:
a)
5
b)
25
c)
125
d)
12.5
|
Vivek Patel answered |
The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse. 
If t1 is the time taken by the planet to travel along ACB and t2 the time along BDA, then- a) t1 = t2
- b)t1 > t2
- c) t1 < t2
- d)Nothing can be concluded
Correct answer is option 'B'. Can you explain this answer?
The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse.
If t1 is the time taken by the planet to travel along ACB and t2 the time along BDA, then
a)
t1 = t2
b)
t1 > t2
c)
t1 < t2
d)
Nothing can be concluded
|
|
Pooja Shah answered |
Here, the angular momentum is conserved i.e. mvr is constant. Where r is the distance from centre of the sun to centre of the planet.
Now, consider one point each from both the mentioned paths. (Shown in fig.)
Applying conservation of angular momentum for these points we get
mv1r1=mv2r2. Simplifying this, v1/v2=r2/r1<1
Time period are given by
t1=L/ v1, t2= L/ v2
Comparing them by taking the ratio
t1/t2=(L/ v1)×(v2/L)
So, t1/t2=v2/v1>1.Thus t1>t2
Now, consider one point each from both the mentioned paths. (Shown in fig.)
Applying conservation of angular momentum for these points we get
mv1r1=mv2r2. Simplifying this, v1/v2=r2/r1<1
Time period are given by
t1=L/ v1, t2= L/ v2
Comparing them by taking the ratio
t1/t2=(L/ v1)×(v2/L)
So, t1/t2=v2/v1>1.Thus t1>t2
In side a hollow spherical shell- a)Everywhere gravitational potential is zero
- b)Everywhere gravitational field is zero
- c) Everywhere gravitational potential is same
- d)Everywhere gravitational field is same
Correct answer is option 'B,C,D'. Can you explain this answer?
In side a hollow spherical shell
a)
Everywhere gravitational potential is zero
b)
Everywhere gravitational field is zero
c)
Everywhere gravitational potential is same
d)
Everywhere gravitational field is same
|
|
Lavanya Menon answered |
The gravitational field inside a uniform spherical shell is 0 from gauss law for gravitation since no mass is enclosed in any Gaussian surface.
Since gravitational potential is given by φ=−∫ gdr, hence, φ=constant since g=0.
Since the gravitational field is 0 everywhere, it is apparently the same everywhere.
Answer is B,C,D.
Since gravitational potential is given by φ=−∫ gdr, hence, φ=constant since g=0.
Since the gravitational field is 0 everywhere, it is apparently the same everywhere.
Answer is B,C,D.
Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.The maximum separation between their centres after their first collision- a)4R
- b)6R
- c)8R
- d)12R
Correct answer is option 'A'. Can you explain this answer?
Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.
The maximum separation between their centres after their first collision
a)
4R
b)
6R
c)
8R
d)
12R
|
|
Kavana Kavana answered |
4
The earth attracts the moon with a gravitational force of 1020 N. Then the moon attracts the earth with the gravitational force of - a)10-20 N
- b)10-2 N
- c)1020 N
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The earth attracts the moon with a gravitational force of 1020 N. Then the moon attracts the earth with the gravitational force of
a)
10-20 N
b)
10-2 N
c)
1020 N
d)
None of the above
|
Ambition Institute answered |
The universal law of gravitation:
- Sir Isaac Newton put forward the universal law of gravitation in 1687 and used it to explain the observed motions of the planets and moons.
- The law states that every particle attracts every other particle in the universe with force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
- Formula, Gravitational force,
where, F is the gravitational force between bodies, m1, and m2 are the masses of the bodies, d is the distance between the centers of two bodies, and G is the universal gravitational constant. - Here, universal gravitational constant, G = 6.67 × 10-11 Nm/kg2
- The Universal Gravitational Law can explain almost anything, right from how an apple falls from a tree to why the moon revolves around the Earth.
Newton's third law of motion
- It states that for every action, there is an equal and opposite reaction.
- Action and reaction always act on two different bodies.
Explanation:
By Newton's Third Law and Newton's Law of Universal gravitation, the gravitational force the Earth exerts on the Moon is exactly the same as the force the Moon exerts on the Earth.
F(Earth) = F(Moon)
Therefore, the moon attracts the earth with a gravitational force of 1020N.
F(Earth) = F(Moon)
Therefore, the moon attracts the earth with a gravitational force of 1020N.
Weight is- a)The force with which an object is pushed towards the centre of the Earth
- b)The force with which an object is pulled towards the centre of the Earth
- c)The force with which an object is pulled towards the centre of the Sun.
- d)The energy with which an object is pulled towards the centre of the Earth
Correct answer is option 'B'. Can you explain this answer?
Weight is
a)
The force with which an object is pushed towards the centre of the Earth
b)
The force with which an object is pulled towards the centre of the Earth
c)
The force with which an object is pulled towards the centre of the Sun.
d)
The energy with which an object is pulled towards the centre of the Earth
|
|
Nandini Patel answered |
Weight is a force caused by gravity. The weight of an object is the gravitational force between the object and the Earth. The more mass the object has the greater its weight will be. Weight is a force, so it's measured in newtons.
What is gravitational potential?- a)It is defined as the work done in taking a unit mass from one point to that point
- b)It is defined as the work done in taking a unit mass from infinity to that point
- c)It is defined as the energy spent in taking a unit mass from infinity to that point.
- d)none
Correct answer is option 'B'. Can you explain this answer?
What is gravitational potential?
a)
It is defined as the work done in taking a unit mass from one point to that point
b)
It is defined as the work done in taking a unit mass from infinity to that point
c)
It is defined as the energy spent in taking a unit mass from infinity to that point.
d)
none
|
Sanjana Bajaj answered |
Understanding Gravitational Potential
Gravitational potential is a fundamental concept in physics, particularly in the study of gravitation and fields. It provides insight into how gravity influences mass in a gravitational field.
Definition of Gravitational Potential
- Gravitational potential at a point in a gravitational field is defined as:
- **The work done in bringing a unit mass from infinity to that point.**
This definition emphasizes that gravitational potential is concerned with the energy required to move an object within a gravitational field.
Why Option B is Correct
- The reason option 'B' is correct can be broken down as follows:
- **Reference Point:** The choice of infinity as a reference point is crucial because at infinity, the gravitational force is effectively zero. This allows us to measure the potential energy relative to a point where gravitational effects can be ignored.
- **Work Done:** The work done against gravitational attraction as an object is brought from infinity to a specified point quantifies the gravitational potential at that point. This work is converted into gravitational potential energy.
Other Options Explained
- **Option A:** Incorrect because it does not specify the reference point (infinity), which is essential for defining gravitational potential.
- **Option C:** While it mentions energy, it inaccurately describes the nature of work done. Gravitational potential is specifically about work done on a unit mass.
- **Option D:** Incorrect as option 'B' is indeed the correct answer.
Conclusion
Understanding gravitational potential is vital for grasping concepts in classical mechanics, as it helps predict the behavior of objects under the influence of gravity.
Gravitational potential is a fundamental concept in physics, particularly in the study of gravitation and fields. It provides insight into how gravity influences mass in a gravitational field.
Definition of Gravitational Potential
- Gravitational potential at a point in a gravitational field is defined as:
- **The work done in bringing a unit mass from infinity to that point.**
This definition emphasizes that gravitational potential is concerned with the energy required to move an object within a gravitational field.
Why Option B is Correct
- The reason option 'B' is correct can be broken down as follows:
- **Reference Point:** The choice of infinity as a reference point is crucial because at infinity, the gravitational force is effectively zero. This allows us to measure the potential energy relative to a point where gravitational effects can be ignored.
- **Work Done:** The work done against gravitational attraction as an object is brought from infinity to a specified point quantifies the gravitational potential at that point. This work is converted into gravitational potential energy.
Other Options Explained
- **Option A:** Incorrect because it does not specify the reference point (infinity), which is essential for defining gravitational potential.
- **Option C:** While it mentions energy, it inaccurately describes the nature of work done. Gravitational potential is specifically about work done on a unit mass.
- **Option D:** Incorrect as option 'B' is indeed the correct answer.
Conclusion
Understanding gravitational potential is vital for grasping concepts in classical mechanics, as it helps predict the behavior of objects under the influence of gravity.
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