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

Subtopic:  Kepler's Laws |
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Suppose the gravitational force varies inversely as the \(n^{th}\) 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. \(R^{\left(\frac{n-2}{2}\right)}\)

Subtopic:  Newton's Law of Gravitation |
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The orbital speed of an artificial satellite very close to the surface of the earth is Vo. Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is 

(a) 4 V0                    (b) 2 V0

(c) 0.5V0                   (d) 4 V0

Subtopic:  Orbital velocity |
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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

Subtopic:  Satellite |
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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. \(\frac{1}{8}~\text{rad/s}\)

Subtopic:  Acceleration due to Gravity |
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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. \(\frac{1}{4}~mgR\)

Subtopic:  Gravitational Potential Energy |
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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. \(8 \pi \sqrt{\frac{R}{g}}\)

Subtopic:  Satellite |
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An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0 . Its potential energy is:

(1) -E0                       

(2) 1.5 E0

(3) 2E0                         

(4) E0

Subtopic:  Gravitational Potential Energy |
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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) 4π2GMT213                      (b) 4πGMR213-R

(c) GMT24π213-R                   (d) GMT24π213+R

Subtopic:  Kepler's Laws |
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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. \(\frac{R}{\left(\frac{gR}{2v^2}-1\right)}\)
2. \(R\left({\frac{gR}{2v^2}-1}\right)\)
3. \(\frac{R}{\left(\frac{2gR}{v^2}-1\right)}\)
4. \(R{\left(\frac{2gR}{v^2}-1\right)}\)

Subtopic:  Gravitational Potential Energy |
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