The question is
I thought of using Kepler's Third Law (a^3=P^2) to solve this problem, but the units didn't seem right. First, I converted the moon's distance in astronomical units by 384,000 km/150,000,000 km, and I got a value of 2.56×10^-3. Cubing this distance, I got 1.68×10^-8. And then I took the square root of this number, which gives you the period of 1.295×10^-4 years. Multiplying that number by 8,760 (365×24), I then got 1.13 days. Is this reasoning correct. Does the Hubble Space Telescope really orbit in 1.13 days? Or am I way off? Thanks.

Indulging in duly licensed professorial behavior - I can't quite tell what orbital radius you took for Hubble. (Hint - it's 400 km away from what?). The units are irrelevant so long as you are consistent, so if you take the Moon's orbital semimajor axis (i.e. radius) in km or Earth radii and period in days or hours, the scaled value for any other Earth orbit will follow with the same units.

a^3=P^2 only when a is in AU, P is in years, and the object is orbiting the sun (M is 1 solar mass). Otherwise, a^3 is only proportional to P^2. So, for objects 1 and 2 orbiting the same body, (a1/a2)^3=(P1/P2)^2. Try that and see what you get.

Dax

I believe that you need to you the distance from the center of the Earth, not just the orbital altitude.

Kepler's third law was made for planetary orbits around the sun, but it can be adapted for other orbits.

Generally, T² = 4pi²/µ a³ where T is orbital period, µ is gravitational parameter (µ=GM), a is semimajor axis.

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Method 1

P = SQRT[ 4 * Pi^2 * r^3 / GM ], where

P = Period (in seconds)
r = Radius of orbit (in meters), equatorial radius of Earth = 6,378,140 m
GM (Earth) = 3.986E+14 m^3/s^2

P = SQRT[ 4 * Pi^2 * (6,378,140+400,000)^3 / 3.986E+14 ] = 5,554 s = 92m:34s


Method 2

P1^2 / R1^3 = P2^2 / R2^3, or
P1 = SQRT[ P2^2 * (R1 / R2)^3 ], where

P1 = Period of #1, i.e Hubble Space Telescope
R1 = Radius of Orbit for #1, for HST R1 = 6,778 km
P2 = Period of #2, for Moon P2 = 27.3 days
R2 = Radius of Orbit for #2, for Moon R2 = 384,000 km

P1 = SQRT[ 27.3^2 * (6,778 / 384,000)^3 ] = 0.0640 days = 92m:10s


The 24 s difference between the two methods is likely due to rounding. Method 1 is the better method because it can be used to calculate the period of a satellite around any body as long a the value of GM is know.

In my previous post I made the following comment:

Although rounding may introduce a small error, I now realize that there is more to it than that. Accurate calculations of a body's orbital period take into account the mass of both bodies. In the case of a satellite, the mass of the satellite to so small in relation to that of the primary (the body being orbited) that the satellite's mass can be ignored. But in the case of the Moon, the mass of the Moon must be considered. Since the orbital period of the Moon is partly determined by it's own mass, the Moon's period cannot be used to determine the period of other orbiting bodies without introducing some error. I believe this is the reason method #2 resulting in a different answer. Method #2 can only be considered an approximation, thus I recommend the use of method #1.

Added: This Web page gives a brief introduction to orbital mechanics.

It seems easier to me to just use the velocity of an object in orbit: v = sqrt(MG/r), where
M is the mass of the Earth: 5.98E24kg
G is the universal graviational constant: 6.67E-11 Nm^2/kg^2
r is the distance from hubble to the center of the Earth in meters. I think it is about 7068 km or 7.068E7m.

This gives a velocity of 7565 m/s. Assuming a circular orbit, a radius of 7068km translates to a circumference of about 43,781,000 m , so it would take 5787 seconds, or 96 minutes to complete one orbit.

If you need more accuracy than that, you would have to take not only the eccentricity of Hubble's orbit into account, but also the gravitational pull of the moon, the oblateness of the earth, etc.

jfribrg,

What you just did is take 2*Pi*r and divided by SQRT[GM/r], or

P = 2 * Pi * r / SQRT[ GM / r ]

= SQRT[ (2 * Pi * r)^2 / (GM / r) ]

= SQRT[ 4 * Pi^2 * r^2 / (GM / r) ]

= SQRT[ 4 * Pi^2 * r^3 / GM ]

which is exactly the equation I gave earlier.

BTW, G and M are often given as the product GM. For Earth GM = 3.986E+14 m^3/s^2

Another little mistake (with big results!) here: I believe you meant to say

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Is this answer pretty close?

I don't understand. What are you trying to ask here?