An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the surface of earth. (R = 6400 km)
(i) Then the distance of satellite from the surface of earth is
Text Solution
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(i) (Easy) Let M and R be the mass and radius of the earth respectively. If m be the mass of
satellite, then escape velocity from earth v e = 
Velocity of satellite v s = 
= 
.........
Further v s = 
= 
∴ 
= 
h = R = 6400 km
(ii) (Easy) T 2 = 
Hence time period of revolution T is
T = 
(Put x = 2R)
∴ T = 
(iii) (Easy) Now total energy at height h = total energy at earth's surface (from principle of
conservation of energy)
∴ 0 – GM 
= 
mv 2 – GM 
or 
mv 2 = 
– 
( h = R)
∴ v = 
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