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`        A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth’s surface. Find (a) its speed in the orbit, (b) its kinetic energy, (c) the time period. Mass of the earth = 6 × 1024 kg.`
3 years ago

Deepak Patra
474 Points
```										Sol. a) V = √GM/r + h = √gr^2/r + h
= √9.8 * (6400 * 10^3)^2/10^6 * (6.4 + 2) = 6.9 * 10^3 m/s = 6.9 km/s
b) K.E. = (1/2) mv^2
= (1/2) 1000 * (47.6 * 10^+6) = 2.38 * 10^10 J
c) P.E. = GMm/-(R + h)
= - 6.67 * 10^-11 * 6 * 10^24 * 10^3/(6400 + 2000) * 10^3 = - 40 * 10^-13/8400 = - 4.76 * 10^10 J
d) T = 2π(r + h)/V = 2 * 3.14 * 8400 * 10^3/6.9 * 10^3 = 76.6 * 10^2 sec = 2.1 hour

```
3 years ago
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