Advertisements
Advertisements
प्रश्न
Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sun
(a) is zero in any small part of the orbit
(b) is zero in some parts of the orbit
(c) is zero in one complete revolution
(d) is zero in no part of the motion.
Advertisements
उत्तर
(b) is zero in some parts of the orbit
(c) is zero in one complete revolution
When a planet is moving in an elliptical orbit, at some point, the line joining the centre of the Sun and the planet is perpendicular to the velocity of the planet. For that instant, work done by the gravitational force on the planet becomes zero. As there is no net increase in the speed of the planet after one complete revolution about the Sun, the work done by the gravitational force on the planet in one complete revolution is zero.
Note:For elliptical orbits angle between force ans velocity is always 90 so there the work done is zero in any small part of the orbit.
APPEARS IN
संबंधित प्रश्न
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
What happens to the force between two objects, if the mass of one object is doubled?
If you compare the gravitational force on the Earth due to the Sun to that due to the Moon, you would find that the Sun’s pull is greater than the Moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the Moon’s pull is greater than the tidal effect of Sun. Why?
State Kepler’s law which is represented by the relation r3 ∝ T2.
Let V and E be the gravitational potential and gravitational field at a distance r from the centre of a uniform spherical shell. Consider the following two statements :
(A) The plot of V against r is discontinuous.
(B) The plot of E against r is discontinuous.
A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth's centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass m when it is at a distance x from the centre of the tunnel.
The law of gravitation gives the gravitational force between :
Explain the following:
People often shake the branches of a tree for getting down its fruits.
A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 ms-2 , find :
the final velocity of the ball on reaching the ground .
Multiple Choice Question. Select the correct option.
The mass of earth is 6 × 1024 kg and radius of earth is 6.4 × 106 m. The magnitude of force between the mass of 1 kg and the earth is:
Explain the difference between g and G.
Is there a gravitational attraction between you and the book? Explain.
Answer the following question.
What are the dimensions of the universal gravitational constant?
Solve the following problem.
Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 1024 kg, R = 6400 km).
To project the rockets which of the following principle(s) is /(are) required?
Law of gravitation gives the gravitational force between
The gravitational force between a hollow spherical shell (of radius R and uniform density) and a point mass is F. Show the nature of F vs r graph where r is the distance of the point from the centre of the hollow spherical shell of uniform density.
The acceleration of the Moon towards the Earth is approximately ______.
Who measured the value of the Universal Gravitational Constant G using the Cavendish balance?
A hollow spherical shell of uniform density exerts what gravitational force on a point mass placed inside it?
