Physics HELP!!!! Attractive Force?

Question

The gravitational force between a pair of neutron stars is 300 N when they are separated by a distance of 10,000 m. What would the attractive force be if the distance was increased by a factor of 3?

I tried 300 N / 3(10,000m)^2 = .000001 N
or do you take half of 10,000 for the radius so it is
300 N / 3(5,000m)^2 = .000004 N
or am I doing it wrong?? Please let me know thanks

Answer 1

http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

The force is inversely proportional to a square of distance, so it would be 9 times smaller.

Answer 2

Fg = (6.67E-11 Mm) / r^2

the relationship between the force and the distance is inverse square.

So if the distance is increased by 3 times, then the force will goes down by 3^2 = 9 times

so 300/9 = 33 1/3N is the force between the two neutron stars when the distance is increased by 3 times

Answer 3

The basic thing to remember is that gravity follows the "inverse square" law. This means that as the distance increases, the gravity gets weaker according to the SQUARE of the distance.

For example:

if you double the distance, gravity decreases to 1/4 (=1/2²) of the original;
if you triple the distance, gravity decreases to 1/9 (=1/3²)
if you multiply distance by 4, gravity decreases to 1/16;
if you multiply distance by 5, gravity decreases to 1/25;
...and so on.

So, in your problem, the distance increases "by a factor of 3". That means the gravity should drop to 1/9 of what it originally was.

As far as your calculation goes, the basic mistake you made was this: You divided 300N by the square of the distance (times 3); but what you SHOULD have done, is divide 300N by the RATIO of the squares of the TWO distances (old distance & new distance). In other words:

Not this: 300N / 3(10,000m)²

But this: 300N / [(30,000m)² / (10,000m)²]

And you can verify with a calculator that (30,000m)² / (10,000m)² = 9. So the force drops by a factor of 9, just like I said.

F = 300N / 9

If you should forget that rule, you can always go back to the original gravity law and work it out that way. Like so:

Original force = F_o = G(m1)(m2)/(10000m)² = 300N

New force = F_n = G(m1)(m2)/(30000m)²

Divide F_n by F_o

F_n / F_o = [G(m1)(m2)/(30000m)²] / [G(m1)(m2)/(10000m)²]

cancelling some stuff out:

F_n / F_o = (10000m)² / (30000m)²

You can verify that (10000m)² / (30000m)² = 1/9; so:

F_n / F_o = 1/9

or, since F_o = 300N:

F_n / 300N = 1/9

F_n = 300N(1/9)

Answer 4

that is not. It purely looks that way. Nicolas has the main appropriate concept. The regulation of gravity is incorrect!!! we've frequently going on this for a while now, yet we nevertheless use it to describe the way products behave by using fact that is stable adequate for ninety 9% of the failings everybody is coping with. Technically you are able to say gravity is a repulsive tension the place area pushes a lot jointly.

Answer 5

300 Newton?????