A planet of mass m = 2.75x 10^24 kg is orbiting in a circular path a star of mas
ID: 1278545 • Letter: A
Question
A planet of mass m = 2.75x 10^24 kg is orbiting in a circular path a star of mass M = 3.15x10^29 kg. The radius of the orbit is R = 8.85x10^7 km. What is the orbital period(in earth days) of the planet Tplanet? Tplanet = ______ days
I have asked this question before but got to many mixed answers, please I really need confirmation, keep an eye on this post I really can't afford to try all the different answers, I lose 5% points each time, thank you for confirming!
A planet of mass m = 2.75x 10^24 kg is orbiting in a circular path a star of mass M = 3.15x10^29 kg. The radius of the orbit is R = 8.85x10^7 km. What is the orbital period(in earth days) of the planet Tplanet? Tplanet = ______ days I have asked this question before but got to many mixed answers, please I really need confirmation, keep an eye on this post I really can't afford to try all the different answers, I lose 5% points each time, thank you for confirming!Explanation / Answer
centripetal force = gravitational force
m?^2R=mMG/R^2
?^2=MG/R^3
T=2?/?
then
T=2?[R^3/(MG)]^1/2 [third law of Kepler]
Solving we get :
T = 2* 3.14 [ (8.85 * 10^10)^3/ (3.15 * 10^29* 6.67384 * 10^-11 ] ^1/2
T = 2*3.14*5.74211298*10^6 = 417.37 days (approx)
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