Determine the binding energy of satellite of mass 1000 kg revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite. [Mass of Earth = 6 x 10^{24} kg, radius of Earth = 6400 km; gravitational constant G = 6.67 x 10-11 Nm^{2} /kg2 ]

#### Solution

Given:- m = 1000 kg, M = 6 x 10^{24} kg, R = 6400 km, G = 6.67 x 10^{-11} N m^{2}/kg^{2}

To find:-

i. Binding Energy (B.E.)

ii. Kinetic Energy (K.E.)

iii. Potential Energy (P.E.)

Formulae:- For satellite very close to earth,

i. B.E. =(1/2)*(GMm/R)

ii. K.E. = B.E.

iii. P.E. = -2K.E.

Calculation: From formula (i),

`B.E=(6.67xx10^-11xx6xx10^24xx1000)/(2xx6.4xx10^6)`

`B.E=(6.67xx6)/(12.8)xx10^10`

= antilog[log 6.67 + log 6 – log 12.8] x 10^{10}

= antilog[0.8241 + 0.7782 – 1.1072] x 10^{10}

= antilog[0.4951] * 10^{10}

= 3.127 x 1010

**∴** B.E. = 3.1265 x 10^{10} J

**The binding energy of the satellite is 3.1265 x 10 ^{10} J.**

From formula (ii),

K.E. = 3.1265 x 10^{10}

**∴** K.E. = 3.1265 x 10^{10} J

**The kinetic energy of the satellite is 3.1265 x 10 ^{10} J.**

From formula (iii),

P.E. = -2(3.1265 x 1010)

**∴** P.E. = -6.2530 x 1010 J

**The potential energy of the satellite is -6.2530 x 10 ^{10} J.**