Asked by Student
A satellite has a mass of 5750 kg and is in a circular orbit 3.6 * 10^5 m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.20 * 10^6 m. What is the true weight of the satellite when it is at rest on the planet's surface?
Ok for this i know we have to use F = ma, and we also use the formula Apparent weight = mg + ma. And that F = (G m1m2)/r^2 plus ac which is v^2/r...now how do i get this started..im actually pretty confused what to do
Ok for this i know we have to use F = ma, and we also use the formula Apparent weight = mg + ma. And that F = (G m1m2)/r^2 plus ac which is v^2/r...now how do i get this started..im actually pretty confused what to do
Answers
Answered by
bobpursley
Yep, you are confused. At orbit, the satellite is weightless, that leads to the conclusion the force of gravity equals the centripetal force.
mv^2/r= GMp m/r^2
note that v= 2PI r/Period.
Put that in for v, divide out the m, divide out a lot of r, and solve for G*Mp.
Now, weight=GMp 5720/(4.2E6)^2
mv^2/r= GMp m/r^2
note that v= 2PI r/Period.
Put that in for v, divide out the m, divide out a lot of r, and solve for G*Mp.
Now, weight=GMp 5720/(4.2E6)^2
Answered by
Student
hold on how do i find the mass of the planet?
Answered by
bobpursley
I suggested you solve for G*Mp
mv^2/r= GMp m/r^2
(2pi*r)^2/r= GMp /r^2
GMp= 4pi^2 r^3
If you want Mp, divide each side by G.
You are given r, the radius to the orbit.
mv^2/r= GMp m/r^2
(2pi*r)^2/r= GMp /r^2
GMp= 4pi^2 r^3
If you want Mp, divide each side by G.
You are given r, the radius to the orbit.
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