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
B
GmM2R
C
GmM3R
D
5GmM6R

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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