Grade 12 Physics - Circular Motion - Universal Gravitation

In summary: C represents some constant.In summary, the question is asking for the magnitude of force between two masses, m1 and m2, when m2 is tripled and the distance between them is halved. The formula FG=(Gm1m2)/r2 is used to calculate the force, with G being a constant of 6.67x10-11. After plugging in the changes to the formula, the resulting simplified equation is Fnew = C x (Gm1m2)/r2, where C represents some constant. The correct answer is 3.1x102N.
  • #1
synchrostarr
1
0

Homework Statement



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?

Homework Equations



FG=(Gm1m2)/r2

G=6.67x10-11

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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  • #2
synchrostarr said:

Homework Statement



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?

Homework Equations



FG=(Gm1m2)/r2

G=6.67x10-11

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
 

Related to Grade 12 Physics - Circular Motion - Universal Gravitation

1. What is circular motion in physics?

Circular motion in physics refers to the movement of an object in a circular path around a fixed point. It is characterized by a constant speed or velocity, and a changing direction.

2. What is the formula for universal gravitation?

The formula for universal gravitation is F = G * (m1 * m2)/r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the two objects.

3. How is circular motion related to universal gravitation?

Circular motion is related to universal gravitation because the centripetal force that keeps an object moving in a circular path is caused by the force of gravity between the two objects. This force of gravity follows the same formula as universal gravitation.

4. What is the difference between centripetal force and centrifugal force?

Centripetal force is the force that keeps an object moving in a circular path, towards the center of the circle. Centrifugal force, on the other hand, is the apparent outward force experienced by an object in circular motion, which is actually a result of inertia.

5. How does the mass and distance between two objects affect the force of gravity between them?

The force of gravity between two objects is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. This means that an increase in either mass or distance will result in a decrease in the force of gravity between the two objects.

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