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The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 × 10³ km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is:

1. 3.2 × 10³ km
2. 1.6 × 10³ km
3. 1.28 × 10⁴ km
4. 6.4 × 10³ km

Correct answer: 3.2 × 10³ km

Solution

E1 = GMm*h/(R(R+h)); E2 = GMm/(2(R+h)). Setting E1=E2 gives h/R = 1/2, so h = R/2 = 6.4x10^3/2 = 3.2x10^3 km.

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