The mean radius of the earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at the earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be -

(1) $R^{2}g/\omega$                   

(2) $R^2{\omega }^2/g$

(3) $Rg/{\omega }^2$                    

(4) $R^{2}g/{\omega }^2$

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. \(T \propto \frac{r^5}{GM}\)
2. \(T \propto \sqrt{\frac{r^3}{GM}}\)
3. \(T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}\)
4. \(T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}\)

Suppose the gravitational force varies inversely as the ${\mathrm{}}^{}$\(n^{th}\) power of distance then the time period of a planet in circular orbit of radius \(R\) around the sun will be proportional to:
1. \(R^{\left(\frac{n+1}{2}\right)}\)
2. \(R^{\left(\frac{n-1}{2}\right)}\)
3. \(R^n\)
4. \(R^{\left(\frac{n-2}{2}\right)}\)

The orbital speed of an artificial satellite very close to the surface of the earth is $V_o$. Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is 

(a) $4 V_0$                    (b) $2 V_0$

(c) $0.5V_0$                   (d) $4 V_0$

The distance of a geostationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to:

(1) 5R                                     (2) 7R

(3) 10R                                   (4) 18R

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be: (\(g= 10~\text{ms}^{-2}\) and the radius of the earth is \(6400\) kms)
1. \(0~\text{rad/s}\)
2. \(\frac{1}{800}~\text{rad/s}\)
3. \(\frac{1}{80}~\text{rad/s}\)
4. \(\frac{1}{8}~\text{rad/s}\)

A body of mass \(m\) is taken from the earth's surface to the height \(h\) equal to the radius of the earth, the increase in potential energy will be:
1. \(mgR\)
2. \(\frac{1}{2}~mgR\)
3. \(2 ~mgR\)
4. \(\frac{1}{4}~mgR\)

Time period of a satellite revolving above Earth’s surface at a height equal to \(R\) (the radius of Earth) will be:
(\(g\) is the acceleration due to gravity at Earth’s surface)
1. \(2 \pi \sqrt{\frac{2 R}{g}}\)
2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)
3. \(2 \pi \sqrt{\frac{R}{g}}\)
4. \(8 \pi \sqrt{\frac{R}{g}}\)

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy ${\mathrm{E}}_0$ . Its potential energy is:

(1) $-{\mathrm{E}}_0$                       

(2) $1.5 {\mathrm{E}}_0$

(3) $2{\mathrm{E}}_0$                         

(4) ${\mathrm{E}}_0$

Given the radius of Earth ‘R’ and length of a day ‘T’, the height of a geostationary satellite is:

[G–Gravitational Constant, M–Mass of Earth] 

(a) ${\left(\frac{4{\mathrm{\pi }}^2\mathrm{GM}}{T^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$                      (b) ${\left(\frac{4\mathrm{\pi GM}}{R^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-R$

(c) ${\left(\frac{{\mathrm{GMT}}^2}{4{\mathrm{\pi }}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-R$                   (d) ${\left(\frac{{\mathrm{GMT}}^2}{4{\mathrm{\pi }}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+R$