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?

2 Answers
Feb 5, 2018

Answer:

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

Explanation:

We need Kepler's third law for this problem, but in the form that is often referred to as "Newton's derivation"

#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.

Here is the solution:

#T^2=((4pi^2)/((6.67xx10^(-11))(5.97xx10^(24))))(1.5xx10^7)^3#

#T^2=3.35xx10^8s^2#

#T=18 292 s#

Divide by 3600 (seconds in each hour) and you get

#T=5.08# hours

Feb 5, 2018

Answer:

The orbital period is #=5.077h#

Explanation:

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 velocity of the satellite is

#v^2=(GM)/r#

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

The orbital period is #=Ts#

The distance is #d=2pir#

Therefore,

#vT=2pir#

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

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

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

#G=6.67*10^-11Nm^2kg^-2#

#M=5.9810^24kg#

So,

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

#=18276.9s#

#=5.077h#