Question

Find the total energy and binding energy of an artificial satellite of mass 1000 kg orbiting at height of 1600 km above the earths surface, [Given : G = 6.67 x 10⁻¹¹ Nm² / kg², R = 6400 km, M = 6400 km, M = 6 x 10²⁴ kg]
(Ans : T.E.= - 2.501 x 10¹⁰J, B.E = 2.501 x 10¹⁰J)

Answer 1

Answer:

T.E.= - 2.501 x 10¹⁰J

B.E.= 2.501 x 10¹⁰J

Explanation:

Since we know that total energy of satellite is the sum of kinetic and potential energy of satellite.

T.E.=K.E.+P.E.

$K.E.=\frac{GMm}{2(R+h)}$..................(1)

$P.E.=\frac{-GMm}{R+h}$....................(2)

So,

$T.E.=\frac{GMm}{2(R+h)}+\frac{-GMm}{R+h}$

$T.E.=\frac{-GMm}{2(R+h)}$.....................(3)

We have G=6.67 x 10⁻¹¹ Nm², R = 6400 km,  M = 6 x 10²⁴ kg, m=1000kg

h=1600km.

put the values in (3) equation

$T.E.=\frac{-6.67\times10^{-11}\times6\times10^{24}\times1000  }{2(6400000+1600000)}$

so T.E.= - 2.501 x 10¹⁰J

Binding energy of satellite=(energy of satellite in infinte orbit-energy of satellite in given orbit)

$B.E.=0-\frac{-GMm}{2(R+h)}$

$B.E.=\frac{GMm}{2(R+h)}$

B.E.= 2.501 x 10¹⁰J

Answer 2

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq