A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?
well, centripetal force= gravityforce from Earth-Gravityforce from Moon
mV^2/Rs=GMem/Rs^2 -GMm*m/(Rm-Rs)^2
where Rs is the orbital distance from earth, Rm is the moon orbital distance.
Now, change to period: 2PI*Rs/Period= V
(2PI)^2*Rs/Period^2=GMe/Rs^2-GMm/(Rm-Rs)^2
Lets look at the period of the moon:
(2PI)^2 Rm/Period^2=GMmMe/Rm^2
or 1/period^2= GMmMe/Rm^3 * 1/(2PI)^2
put that into the equation...
Rs*GMmMe/Rm^3=GMe/Rs^2-GMm/(Rm-Rs)^2
Let Mm/Me = r= ratiomasses
Rs/Rm^3= (1/r)/Rs^2-r/(Rm-Rs)^2
Rs=Rm^3/rRs^2 -r Rm^3/(Rm-Rs)^2
Rs^3=Rm^3/r- rRm^3 (Rs^2/(Rm-Rs)^2)
I think I would turn to a graphical solution at this point.
double check my work. Something is bothering me about this.
http://www.jiskha.com/display.cgi?id=1304796180
The answer I posted previously is wrong. See instead
http://www.ottisoft.com/Activities/Lagrange%20point%20L1.htm
To determine the distance at which the satellite should be placed between the Earth and the Moon, we can use the concept of orbital periods and the Law of Universal Gravitation.
First, let's calculate the orbital radius of the Moon around the Earth. The orbital period of the Moon is 27.3 days. We know that the orbital period (T) can be related to the orbital radius (r) using Kepler's Third Law of Planetary Motion:
T^2 ∝ r^3
Since the Earth and the Moon both orbit around their common center of mass, the distance between them is approximately the same as the Moon's orbital radius. Therefore, we can use the Moon's orbital period to calculate its distance from the Earth.
To simplify the calculations, let's convert the orbital period to seconds. There are 24 hours in a day, and 60 minutes in an hour, and 60 seconds in a minute. So, the orbital period of the Moon is approximately:
27.3 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute = 2,360,320 seconds
Now, we can use this value to calculate the orbital radius of the Moon. Rearranging Kepler's Third Law formula:
r^3 = (T^2 * G * M) / (4π^2)
where G is the gravitational constant (approximately 6.67430 x 10^(-11) m³/(kg·s²)), and M is the mass of the Earth (approximately 5.972 x 10^24 kg).
Plugging in the values:
r^3 = (2,360,320 seconds)^2 * (6.67430 x 10^(-11) m³/(kg·s²)) * (5.972 x 10^24 kg) / (4π^2)
Calculating this expression gives:
r^3 ≈ 2.631 x 10^20 m³
To find r, we need to take the cube root of both sides:
r ≈ (2.631 x 10^20 m³)^(1/3)
≈ 3.844 x 10^8 m
So, the distance between the Earth and the Moon is about 3.844 x 10^8 meters.
Since the satellite should be placed along the same line connecting the centers of mass of the Earth and the Moon, the distance at which the satellite should be placed is also approximately 3.844 x 10^8 meters away from the Earth.