Minimum energy to launch a satellite

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

Give your answer as the coefficient of G m M R \frac{GmM}{R} .


The answer is 0.8333.

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2 solutions

Tapas Mazumdar
Mar 25, 2017

At the desired orbit, we have

Total energy = Potential energy + Kinetic energy = G M m 2 r = G M m 6 R ( where r = R + 2 R = 3 R ) \text{ Total energy } = \text{ Potential energy } + \text{Kinetic energy } = -\dfrac{GMm}{2r} = -\dfrac{GMm}{6R} \qquad \qquad (\small\text{where } r = R+2R = 3R)

At the point of launch near the surface

Total energy = Potential energy = G M m R \text{ Total energy } = \text{ Potential energy } = -\dfrac{GMm}{R}

The amount of energy required will be equal to the amount of work done by an external agent to bring the satellite from the surface of the planet to the desired orbit. Thus

Minimum energy required = Final Total energy Initial Total energy = G M m 6 R + G M m R = 5 G M m 6 R \text{ Minimum energy required } = \text{ Final Total energy } - \text{Initial Total energy } = -\dfrac{GMm}{6R} + \dfrac{GMm}{R} = \dfrac{5GMm}{6R}

Thus the required answer is 5 6 0.8333 \dfrac 56 \approx \boxed{0.8333} .

Lovely Solution! (+1)

Md Zuhair - 3 years, 4 months ago

T. Ef = − GMm 6R T. Ei = − GMm R ∆W = T.Ef – T.Ei = 5GMm 6R

Isn't the name of this question a bit weird.

Spandan Senapati - 4 years, 2 months ago

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Can you write a solution that is helpful to those who cannot solve it? Else I'm inclined to delete this solution to encourage others to contribute a relevant answer.

R G Staff - 4 years, 2 months ago

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sure sir i'll upload pic after maths exam

A Former Brilliant Member - 4 years, 2 months ago

really Mains roll no. are out . i wanted to give a message .

A Former Brilliant Member - 4 years, 2 months ago

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