TA 2019 Physics 1 Question 5: Geosynchronous Orbit

  • Thread starter Thread starter kppc1407
  • Start date Start date
  • Tags Tags
    Orbits
AI Thread Summary
Communications satellites in geosynchronous orbit remain fixed over a point on the equator as the Earth rotates, requiring specific calculations for speed and acceleration. The relevant formulas derive from Newton's second law, F = ma, and involve gravitational force and centripetal acceleration. The equation GMm/r² = mω²r is used, where ω is the Earth's angular speed, allowing for the cancellation of mass (m) and focusing on the radius (r) as the unknown. Understanding the units involved, particularly for the gravitational constant G, is crucial for solving the problem accurately. This discussion emphasizes the importance of correctly applying physics principles to determine satellite motion.
kppc1407
Messages
19
Reaction score
0

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.
 
Physics news on Phys.org
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
 
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top