The change in the potential energy, when a body of mass \(m\) is raised to a height \(nR\) from the Earth's surface is: (\(R\) = Radius of the Earth)
1. \(mgR\left(\frac{n}{n-1}\right)\)
2. \(nmgR\)
3. \(mgR\left(\frac{n^2}{n^2+1}\right)\)
4. \(mgR\left(\frac{n}{n+1}\right)\)

The value of g on the earth's surface is 980 cm/$se{c}^2$ . Its value at a height of 64 km from

the earth's surface is

1.   $960.40 cm/se{c}^2$       

2.   $984.90 cm/se{c}^2$

3.   $982.45 cm/se{c}^2$         

4. 980.45 cm/$se{c}^2$

(Radius of the earth R = 6400 kilometers)     

If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be:

1. g/2                             

2. 4g

3. g/4                            

4. 2g

The masses and radii of the earth and moon are $M_{1, }{R}_1$ and $M_2, R_2$ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is:

1. $2\sqrt{\frac{G}{D}\left(M_1+M_2\right)}$                 

2. $2\sqrt{\frac{2G}{D}(M_1+M_2)}$

3. $2\sqrt{\frac{GM}{D}\left(M_1+M_2\right)}$               

4. $2\sqrt{\frac{GM\left(M_1+M_2\right)}{D\left(R_1+R_2\right)}}$

If the mass of the earth is M, the radius is R and the gravitational constant is G, then work done to take

1 kg mass from earth surface to infinity will be:

1.  $\sqrt{\frac{GM}{2R}}$                      

2. $\frac{GM}{R}$

3.   $\sqrt{\frac{2GM}{R}}$                

4.   $\frac{GM}{2R}$

If $v_e$ and $v_o$ represent the escape velocity and orbital velocity of a satellite corresponding

to a circular orbit of radius R, then 

1. $v_e=v_o$                     

2. $\sqrt{2}{v}_o=v_e$

3. $v_e=v_o/\sqrt{2}$               

4. $v_e$ and $v_o$ are not related

If r represents the radius of the orbit of a satellite of mass m moving around a planet of

mass M, the velocity of the satellite is given by: 

1. $v^2=g\frac{M}{r}$                     

2. $v^2=\frac{GMm}{r}$

3. $v=\frac{GM}{r}$                        

4. $v^2=\frac{GM}{r}$

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of

the earth. R is the radius of the earth and g is acceleration due to gravity at the surface

of the earth. The velocity of the satellite in the orbit is given by:

1. $\frac{g{R}^2}{R+h}$                     

2. gR

3. $\frac{gR}{R+h}$                     

4. $\sqrt{\frac{g{R}^2}{R+h}}$

A satellite which is geostationary in a particular orbit is taken to another orbit. Its

distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time

period in the second orbit is:

1. 4.8 hours                         

2. $48\sqrt{2}$ hours

3. 24 hours                           

4. $24\sqrt{2}$ hours

The ratio of the K.E. required to be given to the satellite to escape earth's gravitational

field to the K.E. required to be given so that the satellite moves in a circular orbit just

above earth atmosphere is:

1. One                           

2. Two

3. Half                           

4. Infinity