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Question:

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 × 10³ km) is E1, and the 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:

3.2 × 10³ km

1.28 × 10⁴ km

6.4 × 10³ km

1.6 × 10³ km

Solution:

Usurface + E1 = Uh
KE of satellite is zero at earth surface
at height h −GMem/Re + E1 = −GMem/(Re + h)
E1 = GMem(1/Re − 1/(Re + h))
E1 = GMem(Re + h − Re)/Re(Re + h)
E1 = GMemh/Re(Re + h)
Gravitational attraction FG = ma
C = mv²/ (Re + h)
E2 ⇒ mv²/(Re + h) = GMem/(Re + h)²
mv² = GMem/(Re + h)
E2 = mv²/2 = GMem/2(Re + h)
E1 = E2
h/Re = 1/2 ⇒ h = Re/2 = 3200 km