According to Kepler’s Law of orbits:
Kepler's Law of Orbits:
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s−1 . If 20 percent of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4 x 1023kg; radius of mars = 3395 km; G = 6.67 x 10-11 N m2 kg-2
Initial velocity of the rocket , v = 2 km/s = 2 x 103 m/s
Mass of mars, M = 6.4 x 1023 kg
Radius of Mars, R = 3395 km = 3.395 x 106 m
universe gravitational constant, G = 6.67 x 10-11 Nm2 kg-2
Initial kinetic energy of the rocket
Initial potential energy of the rocket
Total initial energy
If 20% of initial kinetic energy is lost due to martian atmospheric resistance, then only
80% of its kinetic energy helps in reaching a height.
Total initial energy available
Maximum height reached by the rocket = h
at this height, the velocity and hence,
the kinetic energy of the rocket will become zero
Total energy of the rocket at height h
Applying the law of conservation of energy for the rocket, we can write :
Escape speed from the earth is the
The escape velocity is the minimum velocity required to leave a planet or moon. For a rocket or other object to leave a planet, it must overcome the pull of gravity.
Vescape = √2GM/R
Vescape = 11184 m/sec approximate to 11.2 km/sec
According to Kepler’s Law of areas,
The total energy of a circularly orbiting satellite is
Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the earth’s surface; at the high point, or apogee, it is 4000 km above the earth’s surface. What is the period of the spacecraft’s orbit?
ME = 5.97 x 1024Kg; G= 6.67 x 10-11 Nm2kg-2
According to Kepler’s Law of periods, The _______________ of the time period of revolution of a planet is proportional to the cube of the ___________ of the ellipse traced out by the planet
Kepler's 3rd law is a mathematical formula. It means that if you know the period of a planet's orbit (T = how long it takes the planet to go around the Sun), then you can determine that planet's distance from the Sun (a is the length of the semimajor axis of the planet's orbit)
For a circularly orbiting satellite the potential energy
Energy stored in a capacitor is
As the battery is disconnected, total charge Q is shared equally by two capacitors.
So energy of each capacitor
A ‘central’ force is always directed
The acceleration due to gravity at the North Pole of Neptune is approximately 10.7 m/s2. Neptune has mass 1.0 x 1026 kg and radius 2.5 x 104 km and rotates once around its axis in about 16 h. What is the gravitational force on a 5.0-kg object at the north pole of Neptune?
F = mg
⇒ F = 5×10.7 = 53.5 N
To find the resultant gravitational force acting on the particle m due to a number of masses we need to use:
According to properties of gravitational force, gravitational force between the particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles.
For Geostationary Satellites
The satellite which appear stationary relative to earth, such satellites are called geostationary satellites and will have a period of rotation as the period of rotation of earth i.e 24 Hrs.
Two satellites are in circular orbits around a planet that has radius 9.00 x 106 m . One satellite has mass 68.0 kg, orbital radius 5.00 x 107 m, and orbital speed 4800 m/s. The second satellite has mass 84.0 kg and orbital radius 3.00 x 107 m. What is the orbital speed of this second satellite?
Here v1, = 4800 m/sec; R1 = 5 x 107m, R2 = 3 x 107 m, v2 = ?
The acceleration due to gravity at the North Pole of Neptune is approximately 10.7 x m/s2. Neptune has mass 1.0 x 1026 kg and radius 2.5 x 104 km and rotates once around its axis in about 16 h. What is the apparent weight a 5.0-kg object at Neptune’s equator?
The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is
Deimos, a moon of Mars, is about 12 km in diameter with mass 2.0 x 1015 kg. Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it?
If the radius of earth reduces by 4% and density remains same then escape velocity will
and if density remains constant
So if the radius reduces by 4 % then escape velocity also reduces by 4 %
The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it, is
In Cavendish’s experiment,
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Time period of revolution of earth around sun Te= 1 Year
Time period of revolution of planet around sun,Tp=0.5 Year
Orbital size of earth, re= 1 A.U
Orbital size of the planet,rp=?
Applying Kepler's third law we get:
A particle of mass 3m is located 1.00 m from a particle of mass m. Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero?
Let x be the distance of third particle from the mass 3m. then (1 - x) is the distance will be the distance from mass m. If we require that net force on another mass M be zero, then we must have the following :
Moon has a mass of 7.36 x 1022 kg, and a radius of 1.74 x 106 m. Calculate the acceleration due to gravity on the moon.
If a film of width l is stretched in the longitudinal direction a distance d by force F, surface tension is given by
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth. What is the acceleration due to gravity at the surface of Titania
Data: G = 6.67x10−11 N m2/kg, RE = 6.38 x 106 m, mE = 5.97 x 1024 kg
We know that gravitational acceleration, g = GM/R2
We know that M = Mass of earth /1700 and R = Radius of earth /8
Hence we get g = 64/1700 times the gravitational acceleration of earth
I.e. g = 64/1700 x 9.8
= 0.37 m/s2
Which of the following statements correct?
The acceleration due to gravity due to a body at point P on the surface of earth is
Let the body is placed to point Q at a height h abovr the surface of earth, then acceleration due to gravity:
From above expression we can clearly say that with increase in altitiude h , the value of acceleration due to gravity decreases.
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 108 km.
R = Radius of Orbit of earth = 1.5 x 108 km = 1.5 x 1011m
T = time Period of earth around the sun = 365 Days