# Wikipremed: Keplar's 3rd Law Question

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##### Full Member
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Can someone explain why choice B is false? If a satellite is traveling higher, doesn't it have to cover more distance (resulting in greater speed)?

If anyone can give a simplified version of the explanation for choice A, that would be wonderful also!

#### techfan

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From Newton's Universal law of Gravitation: G*M*m/r^2=F=m*a
Where G is gravitational constant (~6.67*10^-11 N*m^2/kg) M is mass of the larger body (usually Earth for MCAT), m is the mass of the object that is being "attracted" to the larger one, r is the distance between the center of masses of the two objects (r is from the fact that is a radius) and a is the acceleration

Since the object is in circular rotation a=v^2/r (from circular motion fomulae)
G*M*m/r^2=m*v^2/r
m's cancel and 1 r term cancels to get:
G*M/r=v^2

B is incorrect because a higher orbit means the masses are further apart (i.e. r gets larger) and according to the above equation and increase in r would make v^2 (or the speed) smaller. The problem with your reasoning is that you (mistakenly) are assuming that the time it takes to orbit (the period {T}) remains constant for all orbiting bodies. You may be thinking of this like a record player where the tangential velocity (v) is greater on the edges than in the middle, but the angular velocity (omega) is constant. This isn't the case here because mass attraction (gravity) is what is pulling the satellite to the Earth and gravity isn't constant at all distances (in fact it is inversely proportional to the square of the distance {i.e. 1/r^2}). As you move further out the force of gravity decreases.

A is correct because the distance from the center of a circle to the edge is the same (i.e. you only have one value for the radius). This question is trying to get you to understand that for elliptical orbits the distance isn't going to be the same from the center which is why on elliptical orbits the speed is greater when the object gets close to the planet the velocity is high (again gravity is pulling it in).

##### Full Member
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@techfan

I understood up to your last paragraph. Are you saying the radius is constant in a circular orbit (I agree) or that all geosynchronous orbits have the same radii (I don't understand that)?

#### Chrisz

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7+ Year Member

If you have difficulty understanding the explanation. I would like to give you a much easier approach. In the question, it is stated that the orbit is synchronized with the earth's rotation so that the viewver sees it stationary. It is much easier to attack this problem from angular speed. The angular speed of earth's rotation should be equal to the angular speed of the satellite so that the viewer will see it stationary. let's say the angular speed is w. In order to have the same angular speed, the satellite must have the liner speed v=wr. See the linear speed is determined by radius. And, v is determined how far away from the earth. GM/r^2= v^2/r. v=sqr(GM/r)
so sqr(GM/r)=wr. In this equition, G and M are constant, only r is variable, so the angular velocity is determined by radius only

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#### techfan

##### Full Member
10+ Year Member

To answer your question, the orbital radius is the distance from the center of mass the satellite (orbiting object) and the center of mass of the planet.

Geosynchronous orbits (as explained above by chrisz) is an orbit where the viewer can always see the object in the sky at the same position. Since the satellite doesn't move in relationship to the viewer and the Earth spins the orbit has to be circular (since the earth is pretty close to a sphere). Think about a spinning a globe; in order to keep something above Chimborazo, Ecuador (mountain on the equator) you have to make a circular orbit. Even if you move to a different location, say Italy, the orbit is still circular (a smaller circle, but still a circle).

All geosynchronous orbits are circular, think about cutting all the way through a grapefruit, no matter how far up or down you cut it's always a circle on one edge. Thus, no matter where you set the geosynchronous satellite the orbit is going to be circular.

##### Full Member
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@techfan

So is it not possible for there to be an object orbiting the earth (geosynchronously) at like 50 km above position X on the ground (I'm just pulling a random number here) and another object orbiting 51 km (again, above position X)?

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