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A satellite is in a circular parking orbit of radius rp=7337km from the centre of the earth. Determine the velocity increases v2-v1 and v4-v3 necessary to perform a hohmann transfer to a circular orbit with radius equal to the radius of the moon's orbit, 383,000km

Full working and the formula u used would be appreciated

2007-09-29 20:10:03 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

μ = GM = 398600 km^3s^ (-2)
r1 = 7337 km
r2 = 383000 km
2a = r1 + r2 = 390337km
a = 195168.5 km.
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Velocity in the small circular orbit:
V1 = √ [μ/r1] = √ 398600 / 7337
V1 = 7.37 km/s
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Velocity in the larger circular orbit:
V2 = √ [μ/r2] = √ 398600 / 383000
V2 = 1.020 km/s
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VP (velocity at perigee) = V1*√ [r2/a]
VP =7.37*√ [383000/195168.5]
VP = 10.32 km/s
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VA (speed at apogee) = V2* √ [r1/a]
VA = 1.020*√ 7337 / 195168.5
VA = 0.198 km/s
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In the smaller circular orbit the speed is 7.37 km/s, in the larger one 1.020 km/s. In the elliptical orbit in between the speed varies from 10.32 km/s at the perigee to 0.198 km/s at the apogee.
The delta-v's are 10.32 − 7.37 = 2.95 and 1.02 − 0.198 = 0.822 km/s, together 3.72 km/s.

2007-09-30 01:07:57 · answer #1 · answered by Pearlsawme 7 · 0 0

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