Question

A satellite of 600 kg orbits the Earth at a height of 500 km from its surface. Calculate its (i) kinetic energy (ii) potential energy and(iii) total energy ( M = 6 × 1024 kg ; R = 6.4 × 106 m)

Answer 1

radius of earth(R) = 6.4 × 10⁶ m
mass of earth(M) = 6 × 10²⁴ kg
Weight of satellite(m) = 600 kg
height above surface(h) = 500 km = 0.5×10⁶ m
radius of satellite from centre(r) = R+h = 6.9 × 10⁶ m

(i)
Velocity of satellite = $\sqrt{ \frac{GM}{r} }$

⇒ $v= \sqrt{ \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.9 \times 10^6} } = \sqrt{58 \times 10^6}=7.62 \times 10^3\ m/s$

$KE= \frac{1}{2}mv^2= \frac{1}{2} \times 500 \times (7.62 \times 10^3)^2 = 1.45 \times 10^{10} J$

(ii)
$PE =- \frac{GMm}{r} =- \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 500}{6.9 \times 10^6}}=-2.9 \times 10^{10}\ J$

(iii)
$TE=KE+PE=1.45 \times 10^10-2.9 \times 10^{10}=-1.45 \times 10^{10}J$