Gravitational fields

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Homework Statement



What is the centripetal acceleration of a satellite orbiting Saturn at the location exactly one Saturn radius above the surface of Saturn

Homework Equations


Ac=v^2/r
In the previous question found out that the fg at that location is 1/4 of 36.0 N



The Attempt at a Solution


I read somewhere that they made centripetal acceleration equal to Gm/4pi ^2

But why?? How are those two linked?
 

Answers and Replies

  • #2
rude man
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Homework Statement



What is the centripetal acceleration of a satellite orbiting Saturn at the location exactly one Saturn radius above the surface of Saturn

Homework Equations


Ac=v^2/r
In the previous question found out that the fg at that location is 1/4 of 36.0 N



The Attempt at a Solution


I read somewhere that they made centripetal acceleration equal to Gm/4pi ^2

I would not lend much credulity to what you read. The dimensions of Gm/4pi^2 are not the dimensions of acceleration.

Equate gravitational acceleration to centripetal acceleration.
 
  • #3
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I would not lend much credulity to what you read. The dimensions of Gm/4pi^2 are not the dimensions of acceleration.

Equate gravitational acceleration to centripetal acceleration.

The answer is the same as the gravitational intensity which is 2.60N . How does that work??
 
  • #4
rude man
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The answer is the same as the gravitational intensity which is 2.60N . How does that work??

Not too well. Acceleration is not measured in Newtons either ...
 
  • #5
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Then how do u do this question?
 
  • #6
rude man
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I have two masses m1 and m2. What is the (gravitational) force of m1 on m2 (or m2 on m1)?

Picking the force on m2, then use F = ma on m2. Your answer is a. What is gravitational F?
 
  • #7
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The mass being saturn's mass?
 
  • #8
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Well, you have two masses to consider, the satellite's mass and Saturn's mass. After you make some cancellations in your equations, you may not need both values, though.

Anyway, if this satellite is in orbit, it's experiencing a centripetal force. The real question: what force IS PROVIDING this centripetal force? What's the equation for this force that rude man is getting at? Equate this to your centripetal.
 
  • #9
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Well, you have two masses to consider, the satellite's mass and Saturn's mass. After you make some cancellations in your equations, you may not need both values, though.



Anyway, if this satellite is in orbit, it's experiencing a centripetal force. The real question: what force IS PROVIDING this centripetal force? What's the equation for this force that rude man is getting at? Equate this to your centripetal.

Mv^2/r=gm1m2/r^2???
 
  • #11
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How come three different masses?
Oh no, the m in the first equation is the same as m1 in the equation after the equal sign.

So when cancelled would be: v^2/r=Gm/ r^2?
 
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  • #12
rude man
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Oh no, the m in the first equation is the same as m1 in the equation after the equal sign.

So when cancelled would be: v^2/r=Gm/ r^2?

Looking better.
So which is your m? The planet or the satellite? And what is r?
 
  • #13
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Mass would be the mass of Saturn and radius would be the radius of Saturn *2? The thing is , all of this info is not given in the problem, so how are we supposed to know it?
 
  • #14
rude man
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Mass would be the mass of Saturn and r would be the radius of Saturn *2?

Yes.

The thing is , all of this info is not given in the problem, so how are we supposed to know it?

Take a trip to Saturn? Dunno, but you need both the mass and the radius of Saturn.
 
  • #15
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i will try it out and see whether I can it.
 
  • #16
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I tried it out and did not get the answer. Could it be any way that the gravitational field intensity is equal to the centripetal acceleration, since in the book, the answer to both is the same
 
  • #17
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What's the book's answer? I was anticipating the reasoning you said you followed.

Below I pulled Saturn's mass and radius from Google, thought if it's a book problem, I'm sure your text has similar values printed inside.

mv^2/r = GmM/r^2
v^2/r = GM/r^2
where M = mass of saturn = 568.3 x 10^24 kg
r = two saturn radii = 2(58,232,000 m)
a = GM/r^2
a = 2.79 m/s^2
 
  • #18
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What's the book's answer? I was anticipating the reasoning you said you followed.

Below I pulled Saturn's mass and radius from Google, thought if it's a book problem, I'm sure your text has similar values printed inside.

mv^2/r = GmM/r^2
v^2/r = GM/r^2
where M = mass of saturn = 568.3 x 10^24 kg
r = two saturn radii = 2(58,232,000 m)
a = GM/r^2
a = 27.9 m/s^2

The answer is 2.60 m/s^2 which is the same as the gravitational intensity
 
  • #19
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What is this gravitational intensity you keep bringing up?

PS... I edited my post in the time you replied. I had made a calc error. I got a = 2.79 m/s^2, which is pretty close to 2.60m/s^2. I'm wondering if your book has different, perhaps rounded values for Saturn's mass and radius.
 
  • #20
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What is this gravitational intensity you keep bringing up?

PS... I edited my post in the time you replied. I had made a calc error. I got a = 2.79 m/s^2, which is pretty close to 2.60m/s^2. I'm wondering if your book has different, perhaps rounded values for Saturn's mass and radius.

This was a two part question. In the first it asked for gravitational intensity.
 
  • #21
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I think that might be the case. I got the 2.79 m/s^2 too.
 
  • #22
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Oh okay... So gravitational intensity is gravitational force per unit mass... otherwise stated as F/m.

We know gravitational force is GmM/r^2, so therefore gravitational intensity is GM/r^2. Which is... ta dah! The same thing we had before!

Yeah, I'm not sure what digits were used in the prior calculation that got you 2.60m/s^2, but the theory is identical in both approaches. If it was a two stage problem, I'd stick with the 2.6.
 
  • #23
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Oh okay... So gravitational intensity is gravitational force per unit mass... otherwise stated as F/m.

We know gravitational force is GmM/r^2, so therefore gravitational intensity is GM/r^2. Which is... ta dah! The same thing we had before!

Yeah, I'm not sure what digits were used in the prior calculation that got you 2.60m/s^2, but the theory is identical in both approaches. If it was a two stage problem, I'd stick with the 2.6.


Wow! Thank you! I understand where it comes from now!
 
  • #24
rude man
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Post #17 answer is close to yours so the difference is probably due to differences in what you & he/she used for M and r.

What is "gravitational intensity"? I never heard that term used in lieu of gravitational acceleration.
 

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