Asked by Mandy
If the period of the moon were 31 days while its orbit radius 3.8x10^8m, what would the mass of the Earth be?
Answers
Answered by
Damon
31 days (24 hr/day)(3600 s/hr) = 2.68*10^6 seconds
circumference of orbit = 2 pi r = 23.9*10^8
so v, speed = 23.9*10^8/(2.68*10^6
= 8.91*10^2 m/s
Ac = v^2/r = 79.4 * 10^4 / 3.8*10^8
= 20.9 *10^-4 m/s^2
F = m a
G Me m/r^2 = m Ac note m moon cancels
6.67*10^-11 Me/14.4*10^16 =20.9*10^-4
Me = 45.2 *10^23 = 4.52*10^24 Kg
check my arithmetic but that is the right order of magnitude.
circumference of orbit = 2 pi r = 23.9*10^8
so v, speed = 23.9*10^8/(2.68*10^6
= 8.91*10^2 m/s
Ac = v^2/r = 79.4 * 10^4 / 3.8*10^8
= 20.9 *10^-4 m/s^2
F = m a
G Me m/r^2 = m Ac note m moon cancels
6.67*10^-11 Me/14.4*10^16 =20.9*10^-4
Me = 45.2 *10^23 = 4.52*10^24 Kg
check my arithmetic but that is the right order of magnitude.
Answered by
Mandy
Where do you get the 14.4*10^16 from?
Answered by
Damon
orbit radius squared
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