what would its period be? the Moon's orbital period is around the Earth as 27.3 days.
2006-10-07 16:11:40 · 7 answers · asked by princton_girl 2 in Science & Mathematics ➔ Astronomy & Space
And the law is...
P^2 = 4 [pi]^2 a^3 / { G (M1 + M2) }
P = period of the orbit
pi = 3.14159265...
a = semimajor axis of the orbit
G = 6.6728E-11 m^3 kg^(-1) sec^(-2)
M1 = mass of Earth = 5.974E+24 kg
M2 = mass of moon
Plug in the numbers. Remember to use a consistent unit system (MKS).
For the moon that Earth really has...
M2 = 7.345E+22 kg
a = 3.840E+8 meters
P = 2.354E+6 sec = 27.24 days
You can just repeat the math using a different value for the semimajor axis (a).
Hey, breastfed, if you want to solve the three-body problem analytically, then you go right ahead. I'll watch.
When we calculate the position of the Moon, there are also three bodies: Earth, Moon, Sun. We ignore the influence of the sun in the first approximation for the motion of the Moon. We can likewise ignore the Moon in the first approximation when we calculate the motion of the 2nd moon.
2006-10-07 16:37:18 · answer #1 · answered by David S 5 · 2⤊ 0⤋
Tides are the turning out to be and falling of our ocean stages in assessment to land a lot. Tides are brought about with the aid of the gravitational pull of the moon and the sunlight. think of the sunlight to the final suited area of this sq.. think of the moon is at d. human beings at d and b will journey intense tide at the same time as human beings at a and c would have low tide. a d EARTH b sunlight c Now 6 hours previous due while the moon is at c, a and c would have intense tides and d and b would have low tides. Then approximately 6 hours later at b, intense tides lower back at d and b and coffee tides at a and c. Whe the moon is at d or b, the gravity pull of the sunlight and moon each and each artwork mutually. while the moon is at a or c, the moon and sunlight are at suited angles to a minimum of one yet another and the flexibility is vectored, changing the tides to a and c.
2016-12-08 10:26:07 · answer #2 · answered by ? 4 · 0⤊ 0⤋
kris is only partly right. Unfortunately, Kepler's 3rd law won't work, because it only works for the sun and a planet/asteroid. Newton, however, modified Kepler's third law for any two orbiting bodies. Use that one.
2006-10-07 16:31:35 · answer #3 · answered by sparc77 7 · 0⤊ 0⤋
Use Kepler's Third Law to figure it out for yourself.
2006-10-07 16:17:31 · answer #4 · answered by kris 6 · 0⤊ 0⤋
there are 3 bodies people!
moon
earth
and 2nd moon
now calculate correctly
2006-10-07 17:22:07 · answer #5 · answered by breastfed43 3 · 0⤊ 0⤋
Wait a minute while I run and get my slide rule...
2006-10-07 16:19:31 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I don't know.
2006-10-07 16:33:18 · answer #7 · answered by Krissy 6 · 0⤊ 0⤋