What is the minimum energy required to lunch 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 $2 R$ ?

\frac{2 G m M}{3 R}

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\frac{G m M}{2 R}

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\frac{G m M}{3 R}

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\frac{5 G m M}{6 R}

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Solution

The correct option is D

\frac{5 G m M}{6 R}

As we know, Gravitational potential energy = - GM /r
and orbital velocity, v₀ = vGM/R+h
E_f = ½ mv²₀ – GMm/3R = ½ m GM/3R – GMm/3R
= GMm/3R (1/2-1) = - GMm/6R
E_i = - GM/R + K
E_i = E_f
Therefore minimum required energy, K = 5GMm/6R

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What is the minimum energy required to lunch 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?

The correct option is D 5GmM6RAs we know, Gravitational potential energy = - GM /rand orbital velocity, v0 = vGM/R+hEf = ½ mv20 – GMm/3R = ½ m GM/3R – GMm/3R= GMm/3R (1/2-1) = - GMm/6REi = - GM/R + KEi = EfTherefore minimum required energy, K = 5GMm/6R

What is the minimum energy required to lunch 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 $2 R$ ?

The correct option is D $\frac{5 G m M}{6 R}$
As we know, Gravitational potential energy = - GM /r
and orbital velocity, v₀ = vGM/R+h
E_f = ½ mv²₀ – GMm/3R = ½ m GM/3R – GMm/3R
= GMm/3R (1/2-1) = - GMm/6R
E_i = - GM/R + K
E_i = E_f
Therefore minimum required energy, K = 5GMm/6R