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Help Determining Orbital Velocity about the Center of Mass

  • Thread starter franklin37
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  • #1
Hey, so I'm new to Physics Forums as a registered member, but I've been checking this site out for what seems likes years.

Homework Statement



Locate the center of the mass of the Earth-moon system with respect to the center of the Earth, and then find the orbital velocity of the Earth about this center of mass


Homework Equations



R_G = (m1R1 + m2R2) / (m1 + m2)

V_circ = sqrt ( μ / r )

M_E = 5.974 * 10 ^24 kg
M_m = 73.48 * 10^21 kg
Moon semi-major axis of orbit = 384,400 km

The Attempt at a Solution



So, I went ahead and calculated the center of mass of the system. I found that along a straight line connecting the center of the Earth to the center of the Moon, the center of mass was located 4671 km from the center of the Earth.

So now, I need to find the orbital velocity of the Earth about this COM. Using
V_circ = √( μ / r ), I know that I will be able to do so.

I know that in this case, r = 4671 m, as this is the distance from the COM to the center of the Earth. However, I'm having trouble determining the gravitational paramater μ.

In a case like this, is the gravitational parameter of the COM simply μ = G (m_E + m_M)?

This is the first problem that I've had to do that measures velocities in reference to a COM.

Any help would be appreciated
 

Answers and Replies

  • #2
gneill
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Hi Franklin37.

I would be more inclined to use the sidereal period of the Moon's orbit and the circumference of the Earth's circle around the center of mass to find the velocity of the Earth around the center.

The reason is, since two bodies are always diametrically on opposite sides of the center of mass, the line between their centers will have the same angular velocity around the center of mass. That makes their individual "orbital" velocities around that center proportional to R, where R is the distance from the center of mass to the given body (v = ωR). This is not the same relationship as [itex] \sqrt{\mu/R} [/itex].
 
  • #3
thanks a bunch gneill. Your explanation of the fact that the line between their centers will have the same angular velocity about the center of mass was really helpful. So, if the distance of the center of the moon from the COM is roughly 81 times the distance of the the center of the earth from the COM, we can say that the moon's orbital velocity is roughly 81 times that of the Earth's?
 
  • #4
gneill
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20,797
2,777
thanks a bunch gneill. Your explanation of the fact that the line between their centers will have the same angular velocity about the center of mass was really helpful. So, if the distance of the center of the moon from the COM is roughly 81 times the distance of the the center of the earth from the COM, we can say that the moon's orbital velocity is roughly 81 times that of the Earth's?
Yup.
 

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