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:

1.	\(36v_d\)	2.	\(12v_d\)
3.	\(6v_d\)	4.	\(20v_d\)

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

A simple pendulum has a time period ${\mathrm{T}}_1$ when on the earth’s surface, and ${\mathrm{T}}_2$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of ${\mathrm{T}}_2/{\mathrm{T}}_1$ is:

1. 1

2. $\sqrt{2}$

3. 4

4. 2

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

A rocket of mass M is launched vertically from the surface of the earth with an initial speed v. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is   

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

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

3. $\mathrm{R}/\left(\frac{2\mathrm{gR}}{{\mathrm{v}}^2}-1\right)$                         

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

If two planets are at mean distances $d_1$ and $d_2$ from the sun and their frequencies are n₁ and n₂ respectively, then:

1. ${n_1}^2{d_1}^2=n_2{d_2}^2$                                 

2. ${n_2}^2{d_2}^3={n_1}^2{d_1}^3$

3. $n_1{d_1}^2=n_2{d_2}^2$                                   

4. ${n_1}^2{d}_1={n_2}^2{d}_2$