If the acceleration due to gravity at a height \(1\) km above the earth is similar to a depth \(d\) below the surface of the earth, then: 
1. \(d= 0.5\) km
2. \(d=1\) km
3. \(d=1.5\) km
4. \(d=2\) km

Two astronauts are floating in a gravity free space after having lost contact with their spaceship. The two will:

Starting from the centre of the earth having radius R, the variation of g (acceleration  due to gravity) is shown by:

(a)  

(b) 

(c) 

(d) 

A satellite of mass m is orbiting the earth [of radius R] at a height h from its surface. The total energy of the satellite in terms of $g_o$, the value of acceleration due to gravity at the earth's surface is:
1. \(\frac{mg_0R^2}{2(R+h)}\)
2. \(-\frac{mg_0R^2}{2(R+h)}\)
3. \(\frac{2mg_0R^2}{R+h}\)
4. \(-\frac{2mg_0R^2}{R+h}\)

At what height from the surface of earth the gravitation potential and the value of g are $-5.4\times {10}^{7}J$ $k{g}^{-2}$ and $6.0$ $m{s}^{-2}$ respectively? (Take, the radius of earth as 6400 km.)

(a) 1600 km             (b) 1400 km 

(c) 2000 km             (d) 2600 km

The ratio of escape velocity at earth $\left(v_e\right)$ to the escape velocity at a planet $\left(v_p\right)$ whose radius and mean density are twice that of the earth is

Kepler's third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of the average distance r between the sun and planet i.e. T²=Kr^3, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitation, the force of attraction between them is F=GMm/r², here G is gravitational constant. The relation between G and K is described as

(1) GK=4π² 

(2) GMK=4π²
 

(3) K=G
 

(4) K=l/G
 

Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

(1)2.5 R

(2)4.5 R

(3)7.5 R

(4)1.5 R

A remote sensing satellite of the earth revolves in a circular orbit at a height of \(0.25\times 10^{6}\) m above the surface of the earth. If the earth’s radius is \(6.38\times 10^{6}\) m and \(g = 9.8\) ms^-1, then the orbital speed of the satellite is:
1. \(7.76\) kms^-1
2. \(8.56\) kms^-1
3. \(9.13\) kms^-1
4. \(6.67\) kms^-1

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Then,

(1) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

(2) the total mechanical energy of S varies periodically with time

(3) the linear momentum of S remains constant in magnitude

(4) the acceleration of S is always directed towards the centre of the earth