The mean radius of the earth is R, its angular speed on its own axis is ω 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) R2g/ω                   

(2) R2ω2/g

(3) Rg/ω2                    

(4) R2g/ω2

Subtopic:  Orbital velocity |
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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 |
 81%
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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 |
 65%
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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 |
 74%
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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 |
 70%
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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 |
 81%
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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 |
 67%
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