A satellite is in circular orbit at an altitude of 2200 km above the surface of

Question

A satellite is in circular orbit at an altitude of 2200 km above the surface of a nonrotating asteroid with an orbital speed of 3.1 km/s. The minimum speed needed to escape from the surface of the asteroid is 9.8 km/s, and G = 6.67 × 10-11 N · m2/kg2. The mass of the asteroid is closest to
4.0×1023 kg. 7.9×1023 kg. 2.0×1023 kg. 3.0×1023 kg. 6.0×1023 kg. A satellite is in circular orbit at an altitude of 2200 km above the surface of a nonrotating asteroid with an orbital speed of 3.1 km/s. The minimum speed needed to escape from the surface of the asteroid is 9.8 km/s, and G = 6.67 × 10-11 N · m2/kg2. The mass of the asteroid is closest to
4.0×1023 kg. 7.9×1023 kg. 2.0×1023 kg. 3.0×1023 kg. 6.0×1023 kg.
4.0×1023 kg. 7.9×1023 kg. 2.0×1023 kg. 3.0×1023 kg. 6.0×1023 kg.

Explanation / Answer

GMm/(R+h)^2 = m*V^2/(R+h)       V----> orbital velocity

so, GM/(R+h) = V^2       

so, GM = V^2*(R+h)       ...............(1)

 

also,  GMm/R = .5*m*Ve^2         ------> Ve =escape velocity

so, GM = R*Ve^2/2     ............(2)

equating (1) and (2) we get,

R = 550429.6 m

substituting this value of R in either of (1) or (2), we get ,

M =4.0×1023 kg

 

answer :4.0×1023 kg

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