A satellite was placed in an orbit above the earth's atmosphere, at a distance of 15,000 km from the center of the earth. What is the orbital period (in hours) of the satellite?

Question

Charles Aeschliman · Accepted Answer

The orbital period of this satellite is 5.08 hours (5 hours, 5 minutes, 3 seconds to be precise!) .

Kepler's third law—often referred to as "Newton's derivation"—is required for this problem. #T^2=((4pi^2)/(GM))r^3# where G is the gravitational constant #6.67xx10^(-11)# and #M# is the mass of the body that is being orbited (Earth in this case). Also, care must be taken to use base units for #T, r and M#, as the value stated for #G# includes base units only. The answer is as follows: #T^2=((4pi^2)/((6.67xx10^(-11))(5.97xx10^(24))))(1.5xx10^7)^3# #T^2=3.35xx10^8s^2# #T=18 292 s# Subtract 3600 (the number of seconds in an hour) to obtain #T=5.08# hours

Charlotte Aaron · Answer

The orbital period is #=5.077h#

Let the mass of the earth #=Mkg#

Let the mass of the satellite be #=mkg#

Let the radius of the earth be #=Rm#

Let the radius of the satellite's orbit be #=rm#

Let the gravitational constant be #=G#

There are #2# forces acting on the satellite, the gravitational force #F_G# anf the centripetal force #F_C#

#F_G=F_C#

#GMm/r^2=mv^2/r#

The satellite's velocity is

#v^2=(GM)/r#

#v=sqrt((GM)/r)#

The orbital period is #=Ts#

The distance is #d=2pir#

Consequently,

#vT=2pir#

#T=(2pir)/v=2pirsqrt(r/(GM))#

#T=2pisqrt(r^3/(GM))#

#r=15000km=15*10^6m#

G is 6.67 * 10^-11 Nm^2 kg^-2.

#M=5.9810^24kg#

Thus,

#T=2pisqrt((15*10^6)^3/(6.67*10^-11*5.98*10^24))#

#=18276.9s#

#=5.077h#

William Audy · Answer

undefined The orbital period of a satellite in a circular orbit around Earth can be calculated using Kepler's third law of planetary motion. The formula is:

$$ T = 2\pi \sqrt{\frac{r^3}{G M}} $$

Where:
- $ T $ is the orbital period (in seconds).
- $ r $ is the distance from the center of the Earth to the satellite (in meters).
- $ G $ is the gravitational constant ($6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}$).
- $ M $ is the mass of the Earth ($5.972 \times 10^{24} \, \text{kg}$).

Converting the distance from kilometers to meters, $ r = 15,000 \, \text{km} = 15,000,000 \, \text{m} $, we can calculate the orbital period.The orbital period of the satellite is approximately 12.9 hours.