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qUESTIONS

A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

Potential energy of a satellite having mass ‘m’ and rotating at a height of 6.4×10^6 m from the earth surface is

The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance of r2. If v1 and v2 are the linear velocities at these points respectively. Then the ratio v1/v2

The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time to move from A to B then

Kepler’s third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e., T^2 = Kr^3 Here K is constant. If the masses of sun and planet are M and m respectively then as per Newton’s law of gravitation force of attraction between them is F =GMm/r^2 here G is gravitational constant. The relation between G and K is described as

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth is

A body weighs 200N on the surface of the earth. How much will it weigh halfway down to the centre of the earth

The condition for a uniform spherical mass m of radius r to be a black hole is [G= gravitational constant and g= acceleration due to gravity]

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