Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ‘R’ around the sun will be proportional to

Rⁿ
\(R^{\left (\frac{n-1}{2} \right )}\)
\(R^{\left (\frac{n+1}{2} \right )}\)
\(R^{\left (\frac{n-2}{2} \right )}\)

Solution

The time period of an earth satellite in circular orbit is independent of

both the mass and radius of the orbit
radius of its orbit
the mass of the satellite
neither the mass of the satellite nor the radius of its orbit.

Solution

A satellite of mass m revolves around the earth of radius R at a height ‘x’from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

\(\frac{gR^{2}}{R+x}\)
\(\frac{gR}{R-x}\)
gx
\(\left ( \frac{gR^{2}}{R+x} \right )^{1/2}\)

Solution

Energy required to move a body of mass m from an orbit of radius 2R to 3R is

GMm/12R²
GMm/3R²
GMm/8R
GMm/6R

Solution

If the change in the value of g at the height h above the surface of the earth is the same as at a depth ‘x’ below it,then (both x and h being much smaller than the radius of the earth)

x = h
x = 2 h
x = h/2
x = h²

Solution

The time period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4 R is

Solution

The kinetic energy of a satellite inits orbit around the ear this E. What should be the kinetic energy of the satellite so as to enable it to escape from the gravitational pull of the earth?

4 E
2 E
√2 E
E

Solution

If the length of a simple pendulum is increased by 2%, then the time period

increases by 2%
decreases by 2%
increases by 1%
decreases by 1%

Solution

The escape velocity fora body projected vertically up wards from the surface of earth is 11 km/s. If the body is projected at an angle of 45º with the vertical, the escape velocity will be

22 km/s
11 km/s
¹¹⁄_√2 km/s
11√2 km/s

Solution

Since escape velocity \(\left ( V_{e}=\sqrt{2gR_{e}} \right )\) is independent of angle of projection, so it will not change.

The potential energy of a satellite of mass m and revolving at a height R_e above the surface of earth where R_e= radius of earth, is

- m g R_e
\(\frac{-mg\, R_{e}}{2}\)
\(\frac{-mg\, R_{e}}{3}\)
\(\frac{-mg\, R_{e}}{4}\)

Solution

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