Question 227: Gravitation

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 $$2R$$?

Accepted Answer

Energy of the satellite on the surface of the planet
E
i
= K.E + P.E = 0 + $$\left( { - {{GMm} \over R}} ight)$$ = $${ - {{GMm} \over R}}$$
Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v
E
f
= $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} ight)$$
In the orbital of planet, the centripetal force is provided by the gravitational force
$$	herefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} ight)}^2}}}$$
$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$
$$	herefore$$ E
f
= $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} ight)$$
$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$
= $$ - {{GMm} \over {6R}}$$
$$	herefore$$ Minimum energy required required to launch the satellite
= E
f
- E
i
= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} ight)$$
= $${{5GMm} \over {6R}}$$

Answer

$${{5GmM} \over {6R}}$$

Answer

$${{2GmM} \over {3R}}$$

Answer

$${{GmM} \over {2R}}$$

Answer

$${{GmM} \over {3R}}$$

Gravitation

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

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