For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of -

1.	2	2.	\(\sqrt{2}\)
3.	\(1/\sqrt{2}\)	4.	4

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes 1% of its value at the surface is: (Radius of the earth = R)

1. 8R           

2. 9R

3. 10R         

4. 20R

If the density of the earth is increased 4 times and its radius becomes half of what it is, our weight will be:

1. four times the present value.

2.  doubled.

3.  the same.

4.  Halved.

The escape velocity for the Earth is taken \(v_d\). Then, the escape velocity for a planet whose radius is four times and the density is nine times that of the earth, is:

The value of ‘g’ at a particular point is $9.8$ $\mathrm{m}/{\mathrm{s}}^2$. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of ‘g’ at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)

If both the mass and the radius of the earth is decreased by 1%, then the value of the acceleration due to gravity will:

1. decrease by 1%.                 

2. increase by 1%.

3. increase by 2%.                    

4. remain unchanged.

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. $mgR\left(\frac{n}{n+1}\right)$

A satellite whose mass is m, is revolving in a circular orbit of radius r, around the earth of mass M. Time of revolution of the satellite is:

1. $\mathrm{T}\propto \frac{{\mathrm{r}}^5}{\mathrm{GM}}$

2. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^3}{\mathrm{GM}}}$

3. $\mathrm{T}\propto \sqrt{\frac{\mathrm{r}}{{\mathrm{GM}}^2/3}}$

4. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^3}{{\mathrm{GM}}^2/4}}$

Suppose the gravitational force varies inversely as the ${\mathrm{n}}^{\mathrm{th}}$ power of distance then the time period of a planet in circular orbit of radius R around the sun will be proportional to:

1. ${\mathrm{R}}^{\left(\frac{\mathrm{n}+1}{2}\right)}$                 

2. ${\mathrm{R}}^{\left(\frac{\mathrm{n}-1}{2}\right)}$

3. ${\mathrm{R}}^{\mathrm{n}}$                         

4. ${\mathrm{R}}^{\left(\frac{\mathrm{n}-2}{2}\right)}$  

Time period of a satellite revolving above Earth’s surface at a height equal to \(\mathrm{R}\) (the radius of Earth) will be:
(g is the acceleration due to gravity at Earth’s surface)

1. $2\mathrm{\pi }\sqrt{\frac{2\mathrm{R}}{\mathrm{g}}}$

2. $4\sqrt{2}\mathrm{\pi }\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

4. $8\mathrm{\pi }\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$