Answers

2015-09-25T16:29:53+05:30
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
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