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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top