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Question

What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$?


A
2GmM3R
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B
GmM2R
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C
GmM3R
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D
5GmM6R
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Solution

The correct option is D $$\displaystyle \frac{5GmM}{6R}$$

Given that,

Mass of satellite $$=m$$

Mass of planet $$=M$$

Radius $$=R$$

Altitude $$h=2R$$

Now,

The gravitational potential energy

$$P.E=\dfrac{-Gm}{r}$$

Potential energy at altitude $$=\dfrac{GmM}{3R}$$

Orbital velocity $${{v}_{0}}=\sqrt{\dfrac{2GmM}{R+h}}$$


Now, the total energy is

  $$ {{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R} $$

 $$ {{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R} $$

 $$ {{E}_{f}}=\dfrac{GmM}{3R}\left[ \dfrac{1}{2}-1 \right] $$

 $$ {{E}_{f}}=\dfrac{-GmM}{6R} $$

Now, $${{E}_{i}}={{E}_{f}}$$

Now, the minimum required energy

  $$ K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R} $$

 $$ K.E=\dfrac{5GmM}{6R} $$

Hence, the minimum required energy is $$\dfrac{5GmM}{6R}$$ 


Physics

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