NCERT Exemplar: Gravitation | Physics Class 11 - NEET PDF Download

Ans: (d)
If the density of the earth is non-uniform then ‘g’ at different point on the surface of the earth will be different.

If we substitute any value of density, g ≠ 0 at any point.

Ans: (c)
Force of attraction between any two objects obeys the inverse square law as its universal law. The relative motion between Earth, Mercury as observed from Earth will not be circular as the force on Mercury due to the sun is very large than due to Earth and due to the relative motion of Sun and Earth with Mercury.

Ans: (a)
The torque on earth due to gravitational attractive force on earth is zero, because the direction of force (F) (gravitational) and the line joining, the point of application of force (which also is at c.m. of earth) is along the same direction, so angle between is zero or  

Ans: (c)
We know that the total energy of the earth satellite of radius R bounded system is negative
(-GM\ 2 R)
where M is mass of the earth. –
Due to the atmospheric friction (viscous force) acting on satellite, energy decreases continuously, radius of the orbit or height decreases gradually and the satellite spirals down with increasing speed till it bums in the denser layers of the atmosphere.

Ans: (b)
Moon revolves around the earth in a nearly circular orbit. When it is observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So moon is moving under the combined gravitational pull acting on it due to the earth and the sun. Hence, total force on the moon is not central.

Ans: (d)
Like any other planet, asteroids also follow Kepler’s law and move in elliptical orbits around planet due to the central gravitational force acting on them.
small mass of asteroid does not escape the gravitational force due to sun because of large mass of sun and hence, (a) is incorrect. Since gravitational force follow inverse square law, the asteroid will move in circular or elliptical orbits and (b) is incorrect option

Ans:  (d)
Inertial mass:
The mass of an object A can be defined as the mass of the object obtained by measuring their accelerations under equal forces taking mass of B to be 1kg. This is known as inertial mass.
Gravitational mass:
The gravitational force exerted by a massive body on an object is proportional to the mass of the object. If F_A and F_B are the forces of attraction on the two objects due to the earth

Thus

If B is a standard unit mass, by measuring gravitational Force we can obtain the mass of object A which is known as gravitational mass.
By several experiments, it has been concluded that the inertial and gravitational mass are identical.

Ans: (c)
Let F_BC be the force experienced by mass m at point B due to mass M at point C, and F_BA be the force experienced by mass m at point B due to mass 2M at point A.

Suppose AB = x, then
Substituting the values of AB and BC in the above equations (1) and (2)

As F_BA > F_BC, m will move towards point A (position of particle with mass 2M).

Ans: (a, c)
(a) The acceleration due to gravity at an altitude (height),
increasing height (h) decreases the value of gh. option (a) is correct.
(b) Assuming the earth to be a sphere of uniform density, the acceleration due to gravity at a particular depth (d)
Increasing depth (d) decreases the value of gd. Option (b) is incorrect.
(c) If λ is latitude on earth then gλ = g - ω²R cos² λ
As value of cos λ decrease from 0º to 90º ( from 1 to 0). The acceleration due to gravity increases from equator (λ = 0º) to pole (λ = 90º). Option (c) is correct.
(d) The acceleration due to gravity on surface of earth is
So g on earth depends on mass of earth. Option (d) is incorrect.

Ans: (a, b, c)
If the law of gravitation becomes an inverse cube law instead of inverse square law, then for a planet of mass m revolving around the sun of mass M, we can write
(where R is the radius of orbiting planet)
⇒ Orbital Speed
Time period of revolution of a planet

⇒ T² ∝ R⁴
Hence, orbit will not be elliptical.
[for elliptical orbit T² ∝ R³]
The circular orbits of the planets is not possible according to new law of gravitation.
As force
where,
As g' acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic, (as T ∝ R²).
Also, gravitational force inside a spherical shell of uniform density will have some value. So, only option (d) is incorrect.

Ans: (a, c, d)
According to the problem
G' = 10G
From the adjacent diagram
Gravitational field due to the earth

Weight of person = mg' = m x 10 g = 10 mg
force on the man due to the sun  

As r »R (radius of the earth)
⇒ F will be very small.
So, the effect of the sun will be neglected.
Due to this reason gravity pull on the person will increase. Due to it, walking on ground would become more difficult.
Critical velocity, vc is proportional to g, i.e., v^cg
As, g’ > g => v_c‘>v_c
Hence, rain drops will fall much faster.
To overcome the increased gravitational force of the earth, the aeroplanes will have to travel much faster.

Ans: (a, c, d)
Coulomb’s electric force or Electrostatic force of attraction will produce due to opposite charges.
If the sun and the planets carries huge amount of opposite charges, then electrostatic force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added and both are radial in nature.
Both the forces obey inverse square law and are central forces. As both the forces are of same nature, hence all the three Kepler’s laws will be valid.

Ans: (c, d)
 We know that gravitational force exists between the earth and the sun.
F_G = G(M_S x m_e)/ r²
Where M_S is mass of the sun and m_e is mass of the earth.
This provides the necessary centripetal force for the circular orbit of the earth around the sun. As G decreases with time, the gravitational force F_G will become weaker with time. As F_G is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.
Hence, after long time the earth will leave the solar system.

Ans: (a, c, d)

Where M_o = constant
For calculating acceleration due to gravity let
M₁ = Mass of earth = M
M₂ = mass of object
r₁₂ = radius of earth = R
Then magnitude of force between the object and earth is given by

Which can be said as
Where K is a constant
Therefore, acceleration due to gravity can be calculated as

Which depends on mass and hence different for different objects
Therefore (a) is the correct option
since this force follows inverse square law, hence Kepler’s first two are valid but since g is no longer constant, therefore Kepler’s third law (Law of periods) does not remain valid therefore (c) is the correct option
when n is negative than force is given by

Where a = - n
Therefore, force is inversely proportional to mass of body, hence an object lighter than water will experience more force than water and therefore will sink in water
Therefore (a), (c) and (d) are the correct options

Ans: (a, c)
The satellite which appears stationary relative to earth is called the geostationary satellite. It revolves around the earth in the west-east direction with the same angular velocity as done by the earth about its own axis in the west-east direction. A polar satellite revolves around the earth's pole in north-south direction. It is independent of earth's rotation. Option (a), (c) are correct.

Ans: (d)
The center of gravity is based on weight, whereas the center of mass is based on mass. So, when the gravitational field across an object is uniform, the two are identical. However, when the object enters a spatially-varying gravitational field, the COG will move closer to regions of the object in a stronger field, whereas the COM is unmoved.
More practically, the COG is the point over which the object can be perfectly balanced; the net torque due to gravity about that point is zero. In contrast, the COM is the average location of the mass distribution or it is the point where whole mass of the body is supposed to be concentrated. If the object were given some angular momentum, it would spin about the COM.
For small objects, say of sizes less than 100 m placed in uniform gravitational field then centre of mass is very close with the centre of gravity of the body. But when the size of object increases, its weight changes and its CM and CG become far from each other. Like in the case of spherical ball, the CM and the CG are the same, but in case of Mount Everest, its CM lies a bit above its CG.

Ans: Air molecules in the atmosphere are attracted by gravitational force of the earth but they do not fall into earth due to the reason that the molecules in air has some random motion due to temperature, so their resultant motion is not exactly in the vertical downward direction.
But in case of apple, only vertical motion dominates because of being heavier than air molecules. But due to gravity, the density of air is more near to the earth than the density as we go up.

Ans:
Examples of central force:
(i) Gravitational force due to point mass.
(ii) Electrostatic force on the point charge.
Examples of non-central force:
(i) Nuclear force which depends upon the spin of particles.
(ii) Magnetic systems acting between two current carrying loops

Ans: Areal velocity of Mars revolving around the Sun does not change with time according to Kepler’s law, i.e. it is constant with time. Then graph of a real velocity versus time is a straight line parallel to time axis.

Ans: Areal velocity of the earth around the sun is given by

where, L is the angular momentum and M is the mass of the earth.
But angular momentum

Therefore we get,
∴
Therefore, the direction of a real velocity  is in the direction of  
i.e., perpendicular to the plane containing r and v and directed as given by right hand rule. So, areal velocity is normal to the plane containing Earth and Sun as shown in the figure

Ans:  Gravitational force acting between two point masses m₁ and m₂ is given by the relation,
F = G m₁m₂/ r²
It does not depend upon the medium separating the two point masses. Therefore, gravitational force acting between two point masses will remain unaffected when they are dipped in water keeping the separation between them same. Only their apparent weights change, there is no effect on masses.

Ans: Yes, a body can have inertia but no weight. Inertia of a body is related to mass. Where as weight of body is defined as
W = mg
Where
M = mass of body
g = acceleration due to gravity
therefore, weight of the object can be zero when the acceleration due to gravity becomes zero.
Since we know that at the centre of earth, the value of acceleration due to gravity is zero and hence the weight of the object is zero at the centre, but it remains mass(inertia) remains constant.

Ans: The gravitational force does not depend upon the nature of the medium, but the electric force depends on the intervening medium between the charge and the electric field. So we cannot shield a body from the gravitational force.

Ans: Astronaut inside a small spaceship experience a very small negligible constant acceleration and hence astronaut feel weightlessness. If the space station has too much large mass and size then he can only experience acceleration due to gravity in the proximity of moon or on the moon.

Ans:
Given:
Radius of hollow shell = R
Distance of point mass from centre of hollow shell = r
Let the mass of the spherical shell be M and the point mass be m. Then the force between the shell and the point mass is given by,

When point mass is positioned such that it is inside the hollow sphere i.e. 0<r<R, the force is zero as the mass of the hollow sphere is distributed on its surface. When the mass is positioned such that r≤R, the force will be,

Therefore, the graph will be as follows

Ans: Aphelion is the location of earth from where it has the greatest distance from the sun and perihelion is the position from where it has the shortest distance. Now since the angular momentum of earth about the sun is conserved then we have:
mr_ap v_ap = mr_pev_pe
Where m is the mass of the earth, r_ap is the earth - sun distance at the aphelion and r_pe is the earth - sun distance at the perihelion.
∴ r_apv_ap = r_pev_pe

Since r_ap > r_pe, v_pe > v_ap, therefore the velocity at the perihelion is greater than the velocity at the aphelion.

Ans: The orbital plane of a satellite is the plane in which it revolves around the earth. The equatorial plane of earth is the plane which contains the equator.
(a) For a polar satellite, the satellite revolves from north to south and therefore its orbital plane is at 90° to the equatorial plane
(b) For a geostationary satellite whose position with respect to a place on earth is constant i.e. it revolves in the direction of the rotation of earth, the orbital plane is at 0° to the equatorial plane.

Ans:

A day on earth is measured in terms of rotation of earth on its axis with respect to the sun called the mean solar day. When the day is measured with respect to a distant sun, it is called a sidereal day.
Let us look at the given figure, let the sun and distant star’s location be pointed by the long arrow at day 1, where they are directly in line with zenith of the observer on earth. When the earth rotates by 360° the arrow of the distant star points in the same direction along the zenith. However, the sun’s arrow is at an angle from the zenith. Thus, for the sun to be at the zenith in line with the observer, the earth needs to rotate an additional angle. Since the earth moves by 1° around the sun in a day, this additional angle is 1°. Now 360° is equal to 24 hours on earth or 1440 minutes, therefore

Therefore, the distant star will always rise 4 minutes earlier than the sun.

Ans:
Given:
Distance between centres of spheres = 10 x radius
Let the mass of the two spheres be M and their radii be R. Then, the distance between the two will be 10R. let an object of mass m be placed at the midpoint of the line joining the centres of the spheres. Then the force acting on the object due to the two spheres F₁ and F₂ will be,

Therefore, the forces are equal and in opposite directions and therefore the mass is in stable equilibrium.
However, if the mass were to be displaced by say a distance x from the midpoint towards the sphere 2, the forces acting on the object would be


Therefore, F₂’ > F₁’ and therefore the net force acting on the mass will not be zero and equal to (F₂’ - F₁’) towards the sphere 2 and the object would be in unstable equilibrium.

Ans:
Given
Radius of satellite’s orbit = R
(a) Let the mass of earth be M and the mass of the satellite be m. Then the orbital velocity of the satellite will be

then, the kinetic energy of the satellite will be

(b) the potential energy of the satellite is given by

(c) the total energy of the satellite is given by


Therefore, the curves are as follows

Ans: The trajectory or the path followed by a projectile under gravitational force of the earth instead of parabolic will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only the diagram in option (c) will fulfill the requirements.
Important point: In projectile we have taken the value of gravitational acceleration ‘g’ as a constant because up to some distance its variation will be neglected. The trajectory of the particle depends upon the velocity of projection. Depending upon the magnitude and direction of velocity it may be parabolic or elliptical.

Ans:
Given:
Mass of the object = m
Radius of earth = R
Initial height of the object from earth’s centre = R
Final height of the object from earth’s centre = 2R
Let the mass of earth be M. We know that the gravitational potential energy of an object is given by,

Where r is the distance between the object and centre of earth. When the object is at its initial position it is at a distance of R from the centre of the earth. Then its potential energy,

When the object is raised to a height 2R, then the potential energy is given by,

Now the gain in gravitational potential energy is given by

Ans:
Given:
Mass of the ring = M
Layout of the ring = r
Mass of the object = m
Initial distance of the object h = r
Final distance of the object 2h = 2r
To find the gravitational force on the object of mass m due to the ring let us consider an elemental mass on the ring of mass dM. The force due to this elemental mass is then,


where x is the distance of the object from the elemental mass m and

Now at the point P the force dF can be resolved into a vertical component and a horizontal component. If we consider two elemental masses diametrically opposite to each other, then the vertical component due to the two forces cancel each other as they are equal and opposite and the horizontal forces add up. Therefore, we only need to consider the horizontal component of the forces, therefore the net force on the object,

Now from the triangle OPA in the figure we have,

therefore,

When h = r

When h = 2h = 2r

Therefore, the force decreases by a factor of  

Ans: 
Given:
Radius of the orbit of planets = r
Let the mass of the sun be M and the planet be m.
(a) The force on an object revolving around the star is given by


Therefore, linear velocity v is inversely proportional to square root of r and hence when r increases, v decreases.
(b) The angular velocity ω is given by

Therefore, when r increases, ω decreases.
(c) The kinetic energy K is given by,

Therefore, as r increases, K decreases.
(d) The potential energy U is given by,

Therefore, as r increases U becomes less negative or increases.
(e) The total energy E is given by,

Therefore, as r increases E becomes less negative or increases.
(f) The angular momentum L is given by,

Therefore, as r increases, L increases.

Ans:
Given
Side of hexagon = l
The force on the masses will be the resultant of the forces of all the other masses. Let the masses be m each. Now the distance AC is given by the parallelogram law i.e.


Due to symmetry we have AC = AE = 1√3

Now, for AD we have

Now the force on mass at A due to b is

Also
F_ab = F_at
The force on mass A due to C is

Also due to symmetry, we have
F_ac = F_ae
The force on mass at A due to mass at D is

Now, the forces F_ae and F_ac make equal angles with the direction AD viz. 30°, and therefore by parallelogram law of vector addition their resultant is along AD and therefore the magnitude of the resultant,


Similarly, F_ab and F_af also make equal angles with AD viz. 60° and therefore they are also along AD and


Therefore, the net force on mass at A will be
F = F_aec + F_abf + F_ad

Ans: 
(a) For a geostationary satellite, the centripetal force must equal the gravitational force. Let the mass of earth be M and the satellite be m, then

Where r is the radius of the orbit of the satellite and ω is its angular velocity
ω = 2π/T
where T is the time period of revolution, then

Now the height h of the satellite from the surface of the earth,
h = r - R
where R is the radius of earth.
For a geostationary satellite, the time period is 24 hours or 86400 seconds,

(b) A geostationary satellite is visible form the equator in the region shown in the figure having an angle 2θ,

Now from the figure we have

θ = cos^-1[0.152] = 81.29⁰
Now 2θ = 162.58, therefore to cover the earth with such satellites so that from any point on the equator a satellite can be seen,
360/162.58 = 2.21 ≈ 3

Ans:
Given
The eccentricity of earth’s orbit e = 0.0167
Let the earth - sun distance at the aphelion and perihelion be r_a and r_p respectively, and the angular velocity at those points be ω_a and ω_p. Now since the angular momentum of the earth is conserved we have

Now, from the ellipse we have
r_p = a(1-e) and r_a = a(a+e)
Where a is the semi - major axis of the ellipse

Since e = 0.0167
⇒ ω_a/ω_p = 1.069
And since the mean angular velocity is the geometric mean of the two, we have

Now we know that the mean solar day corresponds to 1° on the orbit of earth. A mean solar day has 24 h for which the earth rotates 361°. Thus for aphelion the earth rotates 361.034° and hence the time for the day is 24 hours and 8.1 seconds, and the for the perihelion the earth rotates 359.96° and the time is 23 hours and 52 seconds. Thus the longest day is 8.1 seconds long and the shortest day is 8 seconds shorter. However, the variation of length of day in the year is not explained by this phenomenon as the total time of the day still remains around 24 hours, while the amount of time the sun is visible changes.

Ans:
Given
Radius of earth = R = 6400 km
Aphelion earth - satellite distance r_a = 6R
Perihelion earth - satellite distance r_p = 2R
We know for an ellipse
r_p = a(1-e) and r_a = a(1+e)

Where e is the eccentricity of the ellipse, and the semi major axis.

Now, the angular momentum of the satellite is conserved we then, we have
mr_pv_p=mr_av_a
Where m is the mass of satellite and v_p and v_a are the velocities at the perigee and apogee respectively

Now from conservation of energy at the apogee and perigee
K_P + U_P = K_a + U_a
Where K and U are the kinetic and gravitational potential energy of the satellite

Where M is the mass of earth. Using the relation between v_p and v_a above we have

Putting the values of r_p and r_a we have

∴ v_a = v_P/3 = 2.29 km/s
To transfer the satellite to a circular orbit of radius 6R the satellite’s velocity has to be maintained at a value of