What is the energy required to launch an m kg satellite from earth's surface in a circular orbit at an altitude of 2R? (R=radius of the earth)

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What is the energy required to launch an m kg satellite from earth's surface in a circular orbit at an altitude of 2R? (R=radius of the earth)

$\frac{2}{3}$ $2/3$ mgR

mgR

$\frac{5}{6}$ $5/6$ mgR

$\frac{1}{3}$ $1/3$ mgR

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Answer 1 · 2018-03-10T02:56:37

3.

Explanation:

We need to use Law of conservation of Energy.

$Energy at earth's surface = {PE}_{S} + {KE}_{r} (required for launch)$ $"Energy at earth's surface" = "PE"_S+ "KE"_r "(required for launch)"$

$= - \frac{G M m}{R} + {KE}_{r}$ $= -(GMm)/R +"KE"_r$ .......(1)

$Energy in the orbit at altitude* of 2R = {PE}_{o} + {KE}_{o}^{$}$ $"Energy in the orbit at altitude* of 2R" = "PE"_(o) +"KE"_o^$$

$= - \frac{G M m}{3 R} + \frac{1}{2} m {(\sqrt{\frac{G M}{3 R}})}^{2}$ $= -(GMm)/(3R) + 1/2 m (sqrt((GM)/(3R)))^2$
$= - \frac{G M m}{6 R}$ $= -(GMm)/(6R)$ ......(2)

Equating (1) and (2) and rearranging we get

${KE}_{r} = - \frac{G M m}{6 R} + \frac{G M m}{R}$ $"KE"_r = -(GMm)/(6R)+(GMm)/R$
$\Rightarrow {KE}_{r} = \frac{5 G M m}{6 R}$ $=>"KE"_r = (5GMm)/(6R)$

Writing in terms of acceleration due to gravity $g = \frac{G M}{R^{2}}$ $g=(GM)/R^2$ , we get

$\Rightarrow {KE}_{r} = \frac{5}{6} m g R$ $=>"KE"_r = 5/6mgR$

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

$^{$}$ $"^$$ Calculations for orbital velocity.
Equating Centripetal force to gravitational force in the orbit of radius $r$ $r$

$\frac{m v_{0}^{2}}{r} = \frac{G m M}{r^{2}}$ $(mv_0^2)/r=(GmM)/r^2$
$\Rightarrow v_{0} = \sqrt{\frac{G M}{r}}$ $=>v_0= sqrt((GM)/r)$

*Distances measured from centers of bodies, altitude from surface of earth.

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What is the energy required to launch an m kg satellite from earth's surface in a circular orbit at an altitude of 2R? (R=radius of the earth)

$\frac{2}{3}$ $2/3$ mgR

mgR

$\frac{5}{6}$ $5/6$ mgR

$\frac{1}{3}$ $1/3$ mgR

1 Answer

Explanation:

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1 Answer

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$\frac{2}{3}$ $2/3$ mgR

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