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frozen [14]
3 years ago
9

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

avg_earth = 6.38e6m
Physics
2 answers:
crimeas [40]3 years ago
7 0

Answer:

7020.117 m/s

Explanation:

ivolga24 [154]3 years ago
6 0
<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:  

V=\sqrt{G\frac{M}{R}} (1)  

Where:  

G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}} is the gravity constant

M_{Earth}=5.97{10}^{24}kg the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

R=r_{Earth}+h=8080000m the radius of the orbit (measured from the center of the planet to the satellite).  

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth r_{Earth} and the altitude of the satellite above the Earth's surface h.

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}

V=7020.117\frac{m}{s}  

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Answer:

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Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

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