Question

A 500-kg artificial satellite is revolving around the
earth at a height of 1800 km from the earth. Find out
the potential energy, kinetic energy and total energy
of the satellite. The earth's radius is 6400 km and
g=10m/s^2
Ans. - 2.5 x 1010 J, 1.25 x 1010 J,- 1.25 x 1010 J.

Answer 1

Given data:-

Here,

—› Mass of artificial satellite = m

—› Mass of earth = M (5.972 × 10^24 kg )

—› Gravitational constant = G (6.673 × 10^-11 Nm²/kg²)

—› Radius of earth = R

—› Height of artificial satellite = h

Now, {from que}

—› m = 500 kg

—› R = 6400 km

—› h = 1800 km

Solution:-

We use formulae to find the potential energy, kinetic energy and total energy.

Now, Let,

—› Potential energy = P.E

—› kinetic energy = K.E

—› Total energy = T.E

Here, we calculate,

${\bf{\dashrightarrow{ \pink{ {GMm }}} }}$

${ \tt{ \small{ = 6.673 \times {10}^{ - 11} \times 5.972 \times {10}^{24} \times 500 }}}$

${ \tt{ = 1.9925578 \times {10}^{17} \: N {m}^{2} } \: \: \: \: ....(1)}$

and

${\bf{\dashrightarrow{ \small{ \pink{ { R + h }}} { \tt{ = 6400 + 1800 = 8200}}} \: m \: \: \: \: ......(2)}}$

Now, from eq. ( 1 ) & eq. ( 2 )

${ \dashrightarrow{ \bf{ \pink{ \frac{GMm}{R + h} = \frac{ 1.9925578 \times {10}^{17} }{8200} }}}}$

${ \bf{ \pink{ = 2.429948537 \times {10}^{13} \: Nm }}}$

OR

${ \bf{ \small {\pink{ = 2.43 \times {10}^{13} \: Nm }} \: \: (approx) \: \: \: \: ..... \: (3)}}$

Now, From eq. ( 3 )

${ \bf{ \small{ \dashrightarrow{P.E \: Of \: Satellite{ \: = \pink{ - \frac{GMm}{R + h} }}}}}}$

${ \bf{ \small{ \dashrightarrow{P.E \: Of \: Satellite{ \: = \pink{ - 2.43 \times {10}^{13}J}}}}}}$

Now,

${ \bf{ \small{ \dashrightarrow{K.E \: Of \: Satellite{ \: = \pink{ \frac{GMm}{2(R + h)} }}}}}}$

${ \bf{ \small{ \dashrightarrow{K.E \: Of \: Satellite{ \: = \pink{ \frac{- 2.43 \times {10}^{13}}{2} }}}}}}$

${ \bf{ \small{ \dashrightarrow{K.E \: Of \: Satellite{ \: = \pink{ - 1.215\times {10}^{13} \: J}}}}}}$

Now,

${ \bf{ \small{ \dashrightarrow{T.E \: Of \: Satellite{ \: = \pink{ - \frac{GMm}{2(R + h)} }}}}}}$

${ \bf{ \small{ \dashrightarrow{T.E \: Of \: Satellite{ \: = \pink{ - (\frac{- 2.43 \times {10}^{13}}{2}) }}}}}}$

${ \bf{ \small{ \dashrightarrow{T.E \: Of \: Satellite{ \: = \pink{ 1.215\times {10}^{13} \: J}}}}}}$

Hope it helps you.