Question

An artificial satellite revolves round the Earth at a height 700 kms from the surface. Calculate its :i) Kinetic energyii) Potential energyil) Total energyGiven : Mass of satellite = 150 KgMass of the Earth = 6 x 1024 KgRadius of the Earth = 6,400 Kms.​

Answer 1

We have,

(i) Orbital velocity as,

$\sqrt{ \frac{GM}{r} }$
Therefore,

Kinetic energy=
$\frac{1}{2} m( { \sqrt{ \frac{GM}{r} } )}^{2} \\ \\ = \frac{1}{2} \frac{GMm}{ {r}^{2} } \\ \\ = \frac{1}{2 } \times \frac{6.67 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 150 }{ {(6400 + 700)}^{2} } \\ \\ = 5 . 9 \times {10}^{8} joules$



(ii) Potentials energy is given by,

$- \frac{GMm}{ {r}^{} }$


Therefore the P.E will be,

$- \frac{6.67 \times { 10}^{ - 11} \times 150 \times 6 \times {10}^{24} }{(6400 + 700)} \\ \\ = 84.54 \times {10}^{11} joules$



(iii) Total energy = P.E. + K.E.

$5.9 \times {10}^{8} + 84.59 \times {10}^{11} \\ \\ = 8.4555 \times {10}^{12}$




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