average height of satellite of mass 220kg above earth surface is 640km .it is losing its mechanical energy at a rate of 1.42×10raise to power 5 j in each revolution now answer 1 initial orbital speed 2 height of satellite after 1500 round 3 average regarding force on satellite
Answer (1)
12th Sep, 2020
Hello,
a)Distance from earth's centre = 6400 + 640=7.04 x 10^6m
Now,
mv^2/r= GMm/r^2
V = √GM/r = √56.8 x 10^6
= 7.56 x 10^3 ms ^-1
(B) Total energy after 1500 rounds = Initial energy - Energy loss
= - GMm/2r- 1.42 x 10^5x 1500
=(- 6253-0213) x 10^9
= -6.466 x 10^9J
So, r = - GMm/- 6.466 x 10^9
= 6808 km
Its height above the earth's surface = 6808-6400
= 408 km
(C) Average retarding force, |F |=| dU/dr|
F = Energy loss/Distance
= 1.42 x 10^5/ 2 x 3.14 x 6.924 x 10^3 = 3.3 x 10^-3N
Hope it helps
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