The distance between centre of the earth and moon is 384000 km. If the mass of the earth is $6\times {10}^{24} kg$ and $G=6.66\times {10}^{-11} N{m}^2/k{g}^2$ . The speed of the moon is nearly -

(a) 1 km/sec                          (b) 4 km/sec

(c) 8 km/sec                          (d) 11.2 km/sec

A body is projected vertically upwards from the surface of a planet of radius \(R\) with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is:
1. \(\frac{R}{3}\)
2. \(\frac{R}{2}\)
3. \(\frac{R}{4}\)
4. \(\frac{R}{5}\)

A satellite is launched into a circular orbit of radius \(R\) around the Earth while a second satellite is launched into an orbit of radius \(1.02~\text{R}\). The percentage difference in the time periods of the two satellites is: 

Distance of geostationary satellite from the center of the Earth $(R_e=6400 km)$ in terms of $R_e$ is-

(1) $13.76 R_e$                        

(2) $10.76 R_e$

(3) $6.59 R_e$                            

(4) $2.56 R_e$

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

(1)    $GMm/12R^2$               

(2)    $GMm/3R^2$

(3)     $GMm/8R$                 

(4)     $GMm/6R$

If the radius of the earth contracts by 2% and its mass remains the same, then weight of the body at the earth surface :

(1)  Will decrease                      

(2)  Will increase

(3)   Will remain the same             

(4)  None of these

The kinetic energy needed to project a body of mass m from the earth surface (radius R) to infinity is -

(1)   mgR/2                 

(2)  2 mgR

(3)   mgR                   

(4)   mgR/4

A satellite is to revolve round the earth in a circle of radius 8000 km. The speed at which this satellite be projected into an orbit, will be 
(1) $3 km/s$                     

(2) 16 km/s

(3) 7.15 km/s                   

(4) 8 km/s

If the mass of a body is M on the earth surface, then the mass of the same body on the moon surface is:

(1)  M/6       

(2)  Zero

(3)  M          (4)   None of these

Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to -

1. $\frac{1}{R}$                      

2. $\frac{1}{\sqrt{R}}$

3. R                         

4.  $\frac{1}{R^{3/2}}$