Attraction of two spheres in deep space

  • #1
Feodalherren
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Homework Statement


A solid led sphere of radius 10m has a mass of about 57 million kg. If two of these spheres are floating in deep space with their centers 20m apart, the gravitational attraction between them is only 540N. How large would this gravitational force be if the distance between the centers was tripled?


Homework Equations


F=ma
W=mg
Gm1m2 / d^2



The Attempt at a Solution



This is starting to piss me off.

(G x 57,000,000^2) / 40^2 ≠ 540. It's not even close!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Gravitational force is, as your formula shows, inversely proportional to the square of the distance. If the distance between centers is tripled, the force is multiplied by 1/9.

(where did you get the "40" in the denominator? The given distance between centers is 20 m, not 40. And what did you use for "G".)
 
  • #3
Doc Al
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Gm1m2 / d^2
Use ratios. Compare F1 when the distance = d1 to F2 when the distance = d2 = 3d1.
 
  • #4
Feodalherren
605
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The distance between the spheres was given as 20 but their radius was given as 10m, which makes the distance between the centers a total of 40m. It's still not even close.

Did they pull these numbers out of their a-holes or something!?
 
  • #5
Doc Al
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The distance between the spheres was given as 20 but their radius was given as 10m, which makes the distance between the centers a total of 40m.
You are told that their centers are 20m apart.
 
  • #6
Feodalherren
605
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Woops, you're right. That makes the number even less plausible... Could you calculate it and see if you get 540N?
I must be doing something wrong I get a HUGE number.
 
  • #7
Doc Al
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Woops, you're right. That makes the number even less plausible... Could you calculate it and see if you get 540N?
I must be doing something wrong I get a HUGE number.
Use ratios. If the distance doubles, for example, does the force get bigger or smaller? By what factor?

I will check their numbers out to see if that quoted force makes sense, but you do not need to do that to answer the question.
Edit: Their numbers work out just fine. For the given masses and distance, the force is about 540 N, just like they say.
 
  • #8
Feodalherren
605
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I did use ratios and got 60N. What threw me off was their numbers. They make absolutely no sense.Why would they confuse students by just arbitrarily making up numbers when there's a very specific and easy way to get real numbers!?
 
  • #9
Doc Al
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I did use ratios and got 60N.
Perfectly correct.
What threw me off was their numbers. They make absolutely no sense.Why would they confuse students by just arbitrarily making up numbers when there's a very specific and easy way to get real numbers!?
There's nothing wrong with their numbers. You must have made an error somewhere.
 
  • #10
Feodalherren
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Could you show me how you calculated it, please?

I'm having the same problem with a bunch of questions.
 
  • #11
Doc Al
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Could you show me how you calculated it, please?
You have the equation, just plug in the numbers. What did you use for G?
 
  • #12
Feodalherren
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G= 6.67 x 10^-11

So what I did was (G(57,000,000)^2) / 20^2
 
  • #13
Doc Al
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G= 6.67 x 10^-11
Good.
So what I did was (G(57,000,000)^2) / 20^2
Try it again. You can show your steps if you still can't get it to work.
 
  • #14
Feodalherren
605
6
Actually this time it worked... I'm apparently not calculator savvy.

Thank you very much! Greatly appreciate the help!
 

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