The gravitational field due to a mass distribution is $E=K/x^3$ in the x-direction. (K is a

constant). Taking the gravitational potential to be zero at infinity, its value at a distance x

is

1. K/x                    

2. K/2x

3. $K/x^2$                 

4. K/2$x^2$

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