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lbvjy [14]
3 years ago
7

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

he mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.
Required:
What is the orbital period of this GPS satellite?
Physics
1 answer:
marin [14]3 years ago
4 0

Answer:

T=66262.4s

Explanation:

From the question we are told that:

Altitude A=2.90 *10^7

Mass m=5.97 * 10^{24} kg

Radius r=6.38 *10^6 m.

Generally the equation for Satellite Speed is mathematically given by

V=(\frac{GM}{d} )^{0.5}

V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}

V=3354.83m/s

Therefore

Period T is Given as

T=\frac{2 \pi *a}{V}

T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}

T=66262.4s

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