Question

An artificial satellite revolves round the Earth at a height 700 kms from the surface. Calculate its :
i) Kinetic energy
ii) Potential energy
il) Total energy
Given : Mass of satellite = 150 Kg
Mass of the Earth = 6 x 1024 Kg
Radius of the Earth = 6,400 Kms.
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Answer 1

Explanation:

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

Answer 2

Answer:

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