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Grade 12 Physics - Circular Motion - Universal Gravitation

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synchrostarr
#1
Oct20-13, 10:00 AM
P: 1
1. The problem statement, all variables and given/known data

This question is from the Nelson Grade 12 Physics textbook.

The force of attraction between masses m1 and m2 is 26N in magnitude. What will the magnitude of the force become if m2
is tripled, and the distance between m2 and m1 is halved?

2. Relevant equations

FG=(Gm1m2)/r2

G=6.67x10-11

3. The attempt at a solution

I tried to look at both equations, the original and the altered equation so I had:

26N=(Gm1m2)/r2

and then for the changed equation I had"

FG=(Gm13m2)/(1/2)(r2)

But I was unable to figure out what to do from there.

If it is at all helpful, the answers in the back of my textbook said that the correct answer is 3.1x102N.
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gneill
#2
Oct20-13, 10:37 AM
Mentor
P: 11,867
Quote Quote by synchrostarr View Post
1. The problem statement, all variables and given/known data

This question is from the Nelson Grade 12 Physics textbook.

The force of attraction between masses m1 and m2 is 26N in magnitude. What will the magnitude of the force become if m2
is tripled, and the distance between m2 and m1 is halved?

2. Relevant equations

FG=(Gm1m2)/r2

G=6.67x10-11

3. The attempt at a solution

I tried to look at both equations, the original and the altered equation so I had:

26N=(Gm1m2)/r2

and then for the changed equation I had"

FG=(Gm13m2)/(1/2)(r2)

But I was unable to figure out what to do from there.

If it is at all helpful, the answers in the back of my textbook said that the correct answer is 3.1x102N.
Hi synchrostarr, Welcome to Physics Forums.

You're on the right track. Note that if you halve the distance then the "1/2" applies to r, not to r2.

After you insert the changes into the formula, see if you can't factor them out so you end up with something like:

Fnew = C x (Gm1m2)/r2


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