How Is the Period of a Satellite Calculated?

[KDizzle]

[SOLVED] Period of a satellite

1. Given: G = 6.67 x 10^-11 N m^2/ kg^2
The acceleration of gravity on the surface of a planet of radius R = 4910 km is 11.8 m/s^2. What is the period T of a satellite in circular h = 8445.2 km above the surface? Answer in units of s.



2. F = GMm/r^2
F = ma = mv^2/r
v = sqrt(GM/r)
T = 2pi*r/v



3. i got r = 13355200 m; a = 11.8 m/s^2; that gave me the velocity = 12553.54m/s. i plugged it into the period equation and got T = 6684.425s but its wrong

 

[KDizzle]

nvm i got it

 

[Moetasim]

, the correct answer should be 2.36 hours



I would like to point out that the period of a satellite is directly proportional to the square root of its distance from the center of the planet. In this case, the given satellite is at a distance of 8445.2 km above the surface, which is not the same as the radius of the planet (4910 km). Therefore, the calculated velocity and period will not be accurate.

To obtain the correct answer, we need to use the radius of the planet (4910 km) plus the distance of the satellite from the surface (8445.2 km) to get the total distance (13355.2 km) from the center of the planet. Plugging this value into the period equation, we get a period of approximately 2.36 hours, which is the correct answer.

Furthermore, it is important to note that the given acceleration of gravity on the surface of the planet (11.8 m/s^2) is not relevant to the calculation of the period of the satellite. The period is solely determined by the distance from the center of the planet and the mass of the planet. I hope this helps clarify any confusion.