# Stellite motion problem.

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#1
Hi guys.
How would you show that the speed of the satellite is given by the equation v^2=gr, where g is the gravitational field strength at the orbit.
i tried many different ways such as replace GM/r instead of g in the main formula and also r=v^2/a instead of r but still couldn't get the right answer which leads to v^2=gr .
any help would be much appreciated
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#2
i think i solved it myself using kinetic energy formula and then put instead of v^2, gr .
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5 years ago
#3
You want to post your derivation so that others can see, if they need help?

I have a derivation that uses calculus and vectors and is probably overcomplicating the issue wildly, but I'd like to see how you did it 0
#4
yes sure
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5 years ago
#5
Haven't you just written the kinetic energy formula in terms of g*r and then divided it back out again? If I say a = bc2 and c2 = de, then a = bde = bc2, therefore c2 = de..it shows nothing! it's just basic algebraic manipulation!

Are you trying to derive v2 = gr or are you just trying to show that its true? It seems you're trying to show that it's true but that's definitely not the way to accomplish either of them.

sorry to be an ass about it 0
#6
as the satellite orbits around the earth, it's got kinetic energy which causes the move(which is caused by gravitational strength field of the earth) and one of the factor that has a direct relationship with kinetic energy of the satellite is it's speed therefore kinetic energy of the satellite can help you to express new equation in terms of g and r however it's clearly not something comparable to the mathematic equation you've expressed above so my suggestion to you is try to find a describle reason when you solving a physic problem as well as mathematics to prove it and also don't try to find a link between a mathematic equation and a physic equation because physic needs to be described and also be proved by math .
don't worry about it, we are here to share ideas!
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5 years ago
#7
(Original post by Alen.m)
Hi guys.
How would you show that the speed of the satellite is given by the equation v^2=gr, where g is the gravitational field strength at the orbit.
i tried many different ways such as replace GM/r instead of g in the main formula and also r=v^2/a instead of r but still couldn't get the right answer which leads to v^2=gr .
any help would be much appreciated
I don't think you can use this equation for a satellite though. Since the satellite has mass, it's weight varies inversely proportional to the distance square from the planet and so g is not constant.

The gravitational force (weight) acting on a satellite orbiting a planet is given by the equation: F=GMm/R2 which is equal to the centripetal force holding the satellite in orbit, that is: mv2/R
Therefore, GMm/R2=mv2/R
And so giving v2=GM/R where M is the mass of the planet and R is the Radius of the orbit (the distance from the centre of the satellite to the centre of the earth)
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5 years ago
#8
The thing about physics is that if you take out all meaning of the symbols, it still makes sense mathematically [things still share the same relationships]. Just because you've ascribed meaning to the symbols doesn't mean that they should be treated differently.

I'll try and come up with a new example that is obviously false.
Code:
```my aim is to show that Mass = Volume x height:
M = Vh   Also, F = Ma
so, F = Ma = Vha
since Ma = Vha, M = Vh -- which is what I said before, so it must be true!```
That line of thinking doesn't prove anything. The equation doesn't know what real life application it has, all the equation knows is whether the symbols are sticking to the rules of algebra, which they are. If we try and use this in real life, it will make no sense whatsoever; but that's why we derive the laws using other means -- not flawed algebraic manipulation.

To derive the centripetal acceleration of a body in a circular orbit (or the gravitational field strength), you can make basic assumptions about the orbit and use calculus, like I have in the image below
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5 years ago
#9
...and so g is not constant.
But for a circular orbit [which is what is assumed to begin with], the distance from the centre of gravity of the Earth or other body is constant, because a circle has a constant radius. Since G, M and R are all constant, its acceleration would be too.

If we were discussing elliptical orbits, then you'd be right. Since the distance is not constant, the force, thus the acceleration would vary.

That reminds me, Alen.m I hope you realise that the equation v2=gr is only true for circular orbits, if the satellite was in an elliptical orbit, its speed would change over time
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5 years ago
#10
(Original post by Callum Scott)
But for a circular orbit [which is what is assumed to begin with], the distance from the centre of gravity of the Earth or other body is constant, because a circle has a constant radius. Since G, M and R are all constant, its acceleration would be too.

If we were discussing elliptical orbits, then you'd be right. Since the distance is not constant, the force, thus the acceleration would vary.
Yeah, you are right, but you wouldn't know the acceleration of the satellite and so you cannot use g as the acceleration if you consider g as 9.81m/s2 because that's the acceleration on the surface of the earth. Also, that equation is basically v2=aR where a is the acceleration. If we say v2=gr then we are assuming that the centripetal acceleration is equal to g which could vary depending on the distance of the satellite it's speed
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5 years ago
#11
you cannot use g as the acceleration if you consider g as 9.81m/s2
By g, he and I mean the gravitational field strength [AKA gravitational acceleration], the field strength is measured in newtons per kilogram, aka ms^-2 so it's an acceleration.
We just use the g as the gravitational field strength at the surface a lot in AS physics, but he's using it in the context of anywhere in space. If you look at my derivation, I used a instead of g to avoid the confusion
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5 years ago
#12
(Original post by Callum Scott)
By g, he and I mean the gravitational field strength [AKA gravitational acceleration], the field strength is measured in newtons per kilogram, aka ms^-2 so it's an acceleration.
We just use the g as the gravitational field strength at the surface a lot in AS physics, but he's using it in the context of anywhere in space. If you look at my derivation, I used a instead of g to avoid the confusion
g is always equal to the free fall acceleration at that point is space. That centripetal acceleration equation by itself only holds true for objects in circular motion attached to the centre of the orbit, where there is no force of attraction at work.
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5 years ago
#13
g is always equal to the free fall acceleration at that point is space. That centripetal acceleration equation by itself only holds true for objects in circular motion attached to the centre of the orbit, where there is no force of attraction at work.
I never said that the centripetal acceleration equation holds true for free falling bodies, but what I did say was that a [centripetal acceleration] = g [gravitational field strength] = GM/r2

and by "free fall acceleration", I assume that you mean "the magnitude of the acceleration towards the centre of mass of the Earth", since it doesn't necessarily have to be in free fall.
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5 years ago
#14
(Original post by Callum Scott)
I never said that the centripetal acceleration equation holds true for free falling bodies, but what I did say was that a [centripetal acceleration] = g [gravitational field strength] = GM/r2
Well, I don't know where you think I said that though
Edit: by free fall acceleration I merely meant the same thing because I thought you said g is not equal to a
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5 years ago
#15
Well, I don't know where you think I said that though
"That centripetal acceleration equation by itself only holds true for objects in circular motion attached to the centre of the orbit, where there is no force of attraction at work."

^^ You told me that the equation only holds true for objects in circular orbits, but I never said it didn't -- is the point I was trying to make... but nm; it seems we agree with each other now x'D
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5 years ago
#16
(Original post by Callum Scott)
"That centripetal acceleration equation by itself only holds true for objects in circular motion attached to the centre of the orbit, where there is no force of attraction at work."

^^ You told me that the equation only holds true for objects in circular orbits, but I never said it didn't -- is the point I was trying to make... but nm; it seems we agree with each other now x'D
You didn't but you applied the equation for satellites orbiting the earth which doesn't work
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5 years ago
#17
You didn't but you applied the equation for satellites orbiting the earth which doesn't work
Im sorry, can you quote what I said, or spell it out so a 5 year old could understand it because Im really confused as far as I'm aware, everything i've said so far was correct
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5 years ago
#18
(Original post by Callum Scott)
Im sorry, can you quote what I said, or spell it out so a 5 year old could understand it because Im really confused as far as I'm aware, everything i've said so far was correct
We cannot use the equation v2=gR for satellites orbiting a planet even if you happen yo have the centripetal acceleration of the satellites. This is because the proportionality of R (distance of the orbit) with respect to v2 (tangential speed of the satellites) is not correct. The equation implies that as the distance R increases, the velocity of the satellite must increase which is not correct considering that also g varies as R varies.
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5 years ago
#19
We cannot use the equation v2=gR for satellites orbiting a planet even if you happen yo have the centripetal acceleration of the satellites. This is because the proportionality of R (distance of the orbit) with respect to v2 (tangential speed of the satellites) is not correct. The equation implies that as the distance R increases, the velocity of the satellite must increase which is not correct considering that also g varies as R varies.
What I was saying was that the symbol g, in this case, is synonymous with a -- acceleration. We used them in the same sense [NOT MEANING 9.81ms-2]. g represents the gravitational field strength and is synonymous with the magnitude of the acceleration.

the equation v2=gr, in my eyes, was synonymous with v2=ar. Because g ≠ 9.81ms-2.
g = a = GM/r2

v2=gr therefore v2=GM/r, as the radius increases, the velocity decreases!
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5 years ago
#20
(Original post by Callum Scott)
What I was saying was that the symbol g, in this case, is synonymous with a -- acceleration. We used them in the same sense [NOT MEANING 9.81ms-2]. g represents the gravitational field strength and is synonymous with the magnitude of the acceleration.

the equation v2=gr, in my eyes, was synonymous with v2=ar. Because g ≠ 9.81ms-2.
g = a = GM/r2

v2=gr therefore v2=GM/r, as the radius increases, the velocity decreases!
You are right, but I don't understand why you would use v2=aR instead of V2=GM/R, the equation (v2=aR) would assume that a is constant at any given distance
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