# Orbital speed, Earth vs Moon

1. Jan 1, 2010

### robotnut

Just can't get my head around this one. If at 100Km high above earth the orbital speed to maintain altitude is X, then what would the speed be for the same object around the moon?
Since gravity of the moon is smaller then earth's? I assume the orbital speed would need to be less?

2. Jan 1, 2010

### amaresh92

orbital velocity might be less as the orbital velocity depends on the radius of the orbit in which they move and the mass of moon is less than earth .

3. Jan 2, 2010

### hisham.i

Considering this mass orbiting around the earth or around the moon means its covering a uniform circular motion.
Studying the sum of forces =0, where the forces are:
F=(G.M.m)/d^2 :Gravitational force done by the earth on the object which is the centripetal force.
F1 :Centrifugal force which is also equal in magnitude to the centripetal force, but opposite direction.
applying newtons 2nd law and knowing that the centripetal force is also given by the following fomula F=(m.V^2)/d
then [(G.M.m)/d^2]=[(m.V^2)/d]
calculate v and you will notice that it will be less than the speed on the earth.

4. Jan 2, 2010

### LURCH

To avoid any confusion, we should all agree that these altitudes are above the center of the earth and Moon, respectively; not above their surfaces.

5. Jan 2, 2010

### D H

Staff Emeritus
That is contrary to the meaning of the term "altitude", which invariably means height above some reference surface.

6. Jan 2, 2010

### arunma

This seems like a straightforward question. For a circular orbit (which is the only kind where you get a constant speed), we just set the gravitational field equal to the centripetal acceleration

$$\dfrac{v^2}{r} = \dfrac{GM}{r^2}$$

$$v = \sqrt{\dfrac{GM}{r}}$$

So if you're at some altitude h, just set the distance from the center equal to the radius of the planet R, plus the altitude

$$v = \sqrt{\dfrac{GM}{R + h}}$$

The required speed is independent of the mass (this is for precisely the same reason as why all objects fall at the same acceleration without air resistance). Just set h to 100 km, and you can calculate the orbital speed for the earth and the moon by substituting the appropriate values for the radius and mass of the earth and the moon.

7. Jan 2, 2010

### diazona

of the orbiting body, of course (I know that's what you meant, just thought a clarification couldn't hurt)

If I remember correctly, the moon's mass is 1/6 that of Earth, and it's radius is about 1/3 of the Earth's, so I guess the orbital velocity at a given (small) altitude is reduced by about a factor of 2...

8. Jan 2, 2010

### D H

Staff Emeritus
The Moon's surface gravity is about 1/6 that of the Earth. The Moon's mass however is about 1/81 Earth masses.

9. Jan 2, 2010

### arunma

You're quite right, thanks!

10. Jan 3, 2010

### LURCH

Yes, I know. But what I'm asking is that we all refer to the distance above center of mass, because it removes an unnecessary complication, and avoids the need for secondary calculations such as...

Distance above the surface does not effect orbital dynamics, and I believe this would help the OP gain the understanding he seeks.

11. Jan 3, 2010

### D H

Staff Emeritus
Whatever makes you think that? Distance above the surface most definitely does affect orbital dynamics. Moreover, distance below the surface most definitely hinders orbital mechanics. A satellite could not orbit the Earth at the orbital radius corresponding to an orbital altitude of 100 km above the surface of the Moon because that orbital radius is well inside the Earth. A satellite can however orbit 200 km above the surface of the Earth or the Moon.

The problem with diazona's analysis was incorrect values for the Moon's mass and radius. So, let's do it correctly.

The orbital velocity for a circular orbit at altitude h is

$$v=\sqrt{\frac{GM}{R+h}}$$

The ratio of the orbital velocities for orbits at the same altitude about the Earth and Moon is thus

$$\frac{v_e}{v_m} = \sqrt{\frac{M_e}{M_m}\,\frac{R_m+h}{R_e+h}} \approx \sqrt{\frac{Me}{Mm}\,\frac{R_m}{R_e}}\,\left(1+\frac h 2\,\frac{R_e-R_m}{R_eR_m}\right)$$

Using the correct numbers, Mm/Me=0.0123 and Rm/Re=0.273, yields ve/vm=4.71 for h=0, increasing as altitude increases. The ratio is 5.15 for h=500 km, at which point the approximation is still valid to within about 1%.