TA 2019 Physics 1 Question 5: Geosynchronous Orbit

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SUMMARY

The discussion focuses on calculating the speed and acceleration of a satellite in a geosynchronous orbit, specifically using the parameters of Earth's radius (6.37 x 106 m) and the altitude of the orbit (3.58 x 107 m). The relevant equations include Newton's law of gravitation and centripetal acceleration, expressed as GMm/r2 = mωe2r, where ωe represents the Earth's angular speed. The solution emphasizes the importance of using MKS (SI) units, specifically Newtons (N), kilograms (kg), and meters (m), for accurate calculations.

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Homework Statement



Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the Earth rotates. The radius of the Earth is 6.37 x 106m, and the altitude of a geosynchronous orbit is 3.58 x 107m. What are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?

Homework Equations



M (mass of the earth) = 5.98 x 1024
G (gravitational constant) = 6.67 x10-11
GMm = mv2
r2... r

The Attempt at a Solution



I understand the problem and how to get the answer. I just do not understand where the units go on the above formula and where it is derived from.
 
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Use MKS (SI) units, so units used are N, kg, and m. You may need to look up the units that go with G; it should be in your textbook.

The formula is a result of Newton's 2nd law, F = ma.

Hope that helps.
 
kppc1407 said:
I understand the problem and how to get the answer. I just do not understand where the units go on the above formula and where it is derived from.
Just elaborating on what Redbelly has said, the correct equation is better written as:

\frac{GMm}{r^2} = m\omega_{e}^2r

where \omega_e is the rotational (angular) speed of the earth. Since we know the rotational speed of the Earth and m cancels out, the only unknown is r.

The left side is the force of gravity on a body of mass m at a distance r from the Earth's center. The right side is mass x the (centripetal) acceleration on the body. So the equation is simply an application of F=ma, as Redbelly has stated.

AM
 
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