A particle of mass \(m\) is thrown upwards from the surface of the earth, with a velocity \(u.\) The mass and the radius of the earth are, respectively, \(M\) and \(R.\) \(G\) is the gravitational constant and \(g\) is the acceleration due to gravity on the surface of the earth. The minimum value of \(u\) so that the particle does not return back to earth is:

1. \(\sqrt{\dfrac{2 {GM}}{{R}^2}} \)

2. \(\sqrt{\dfrac{2 {GM}}{{R}}} \)

3.\(\sqrt{\dfrac{2 {gM}}{{R}^2}} \)

4. \(\sqrt{ {2gR^2}}\)

Subtopic:  Escape velocity |
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AIPMT - 2011
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For a satellite, the escape velocity is \(11.2\) km/s from the surface of the earth. If the satellite is launched at an angle of \(60^\circ\) with the vertical, then the escape velocity will be:
1. \(11.2 ~\cos60^\circ\) km/s 2. \(11.2 ~\sin60^\circ\) km/s
3. \(11.2\) km/s 4. \(11.2 ~\tan60^\circ\) km/s
Subtopic:  Escape velocity |
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If the radius of a planet is \(R\) and its density is \(\rho, \) the escape velocity from its surface will be

1. \(V_{e} \propto p R\)                         

2. \(V_{e} \propto R \sqrt{\rho}\)

3. \(V_{e} \propto \dfrac{\sqrt{p}}{R}\)                        

4. \(V_{e} \propto \dfrac{1}{\sqrt{p} R}\)

 

Subtopic:  Escape velocity |
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If mass of a planet is \(9\) times that of the earth and radius is \(2\) times that of the earth, then the escape speed from this planet is:
(\(v_{e}\)is escape speed from the Earth.)
1. \(\dfrac{v_e}{\sqrt2}\)

2. \(\dfrac{v_e}{2\sqrt2}\)

3. \(\dfrac{3v_e}{\sqrt2}\)

4. \(\dfrac{v_e}{2}\)
Subtopic:  Escape velocity |
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