What Are the Correct Calculations for a Geosynchronous Satellite's Orbit?

AI Thread Summary
The discussion focuses on calculating the force acting on a geosynchronous satellite and determining the mass of the planet it orbits. The satellite's orbital speed was calculated to be approximately 14617.1 m/s, leading to a planet mass of about 6.43615 × 10^26 kg. A participant suggested using the formula F = m(v^2/r) for calculating the force, which resulted in a value of approximately 1.016 kN, significantly lower than the initial calculation. The use of 86400 seconds, representing 24 hours, in the force equation was clarified as necessary for accurate calculations. The final consensus indicates that the recalculated force is around 1016.21 N, confirming the need for careful verification of calculations.
jj8890
Messages
43
Reaction score
0
[SOLVED] Circular orbit question...please check answers

Homework Statement


Given: G = 6.67259 × 10^-11 N m^2/kg^2 .
A 956 kg geosynchronous satellite orbits a planet similar to Earth at a radius 201000 km
from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satelliteis 24 h .

What is the force acting on this satellite? (Newtons)
What is the mass of this planet? (kgs)

I just need help checking my answers and make sure that I am using the correct equations. I would appreciate the help.

Homework Equations


v= (2*pi*R)/T
v= (Sqrt(G * Mcentral))/R; G=6.67259 *10^-11
F=(GmM)/r^2

The Attempt at a Solution


v= (2*pi*R)/T = (2*pi*201,000,000)/86400 = 14617.1 m/s

v= (Sqrt(G * Mcentral)); (v^2 *r)/G=Mcentral
Mcentral= (14617.1^2 * 201,000,000)/(6.67259 *10^-11)= 6.43615 * 10^26 kg

F=(GmM)/r^2
F= [(6.67259*10^-11) * (956) * (6.43615 *10^26)]/(86400^2)=5.49985 *10^9 N
 
Physics news on Phys.org
Where did the 86400 come from in your force equation?

Your speed seems fine, and I would probably use F=m\frac{v^2}{r} for the force due to circular motion, which gives 1.016 kN. The mass of the planet is then 643*10^24 kg, as you have.

Recheck your force.
 
The 86400 = 24 hrs in seconds, the period. I also thought that the force was high...I'll recalculate.
 
I also got 1016.21 N when recalculated...thanks
 
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 .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top