**Answer the following question in detail.**

Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.

Given: G = 6.67 × 10^{-11} Nm^{2}/kg^{2}

R = 6400 km, M = 6 × 10^{24} kg

#### Solution

**Given:** m = 2000 kg, h = 3600 km = 3.6 × 10^{6} m,

G = 6.67 × 10^{-11} Nm^{2}/kg^{2},

R = 6400 km = 6.4 × 10^{6} m,

M = 6 × 10^{24} kg

**To find: **

1. Kinetic energy (K.E.)

2. Potential energy (P.E.)

3. Total energy (T.E.)

4. Binding energy (B.E.)

**Formulae: **

1. K.E. = `"GMm"/(2("R + h"))`

2. P.E. = `- "GMm"/("R + h")` = - 2(K.E.)

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

4. B.E. = –T.E.

**Calculation:**

From formula (i),

K.E. = `(6.67 xx 10^-11 xx 6 xx 10^24 xx 2 xx 10^3)/(2 xx [(6.4 xx 10^6) + (3.6 xx 10^6)])`

`= (6.67 xx 6 xx 10^16)/10^7`

= 40.02 × 10^{9} J

From formula (ii),

P.E. = –2 × 40.02 × 10^{9 }= - 80.04 × 10^{9} J

From formula (iii),

T.E. = (40.02 × 10^{9}) + (–80.02 × 10^{9}) = - 40.02 × 10^{9} J

From formula (iv),

B.E. = – (–40.02 × 10^{9}) = 40.02 × 10^{9} J

Kinetic energy of the satellite is 40.02 × 10^{9} J potential energy is –80.04 × 10^{9} J, total energy is -40.02 × 10^{9} J and binding energy is 40.02 × 10^{9} J.