Question

newton's universal law of gravitation asserts that two point masses m1 and m2 seperated by a distance s, attract one another with a force F=(km1m2)/s^2 , k being a universal constant.

Answer 1

wats the main objective of this question?

Answer 2

sorry never got to finish
if s=16 meter, use differentials to estimate the change in s that will decrease the force F by 10%

Answer 3

So, find ds : dF = 0.1F?

Answer 4

sorry super tired so use related rates? dF is 0.1F and differentiate and solve for ds?

Answer 5

That's what I'm thinking, but I'm not sure if I'm setting it up right. I scribbled out some stuff and got a negative value for ds, and that can't be right because decreasing s will increase F...

Answer 6

I think I need to put in dF = -0.1F . . .
maybe. I'm a bit tired too. X-D

Answer 7

ya a neg 0.1 makes more sense to me too

Answer 8

I don't even think this is a related rates thing. It's more like using partials to find marginal cost or %error or something like that.

Answer 9

ok good cuz i was stuck thinking how to take the derivative of km1m2..

Answer 10

I'm assuming km1m2 stays constant.

Answer 11

I'm getting what I think is a reasonable answer. I'm going to test it with some arbitrary values to see if it's in the ballpark.

Answer 12

I let C=km1m2, so F=C*s^-2
then dF/ds = -2C*s^-3
etc.

Answer 13

my first attempt was F-(F*0.1)=(km1m2)/s^2
so if i substitute C in I getcha ok thanks!

Answer 14

Ok, this does seem like the right way to go. I tested it against the actual difference and the estimate is accurate to 0.8%
Solve for ds and use dF=-0.1F.

Answer 15

ok i feel pretty dumb in asking but am i completely off track thinking of -0.1F=-2C*s^-3 ?

Answer 16

There does seem to be something missing...

Answer 17

lol please don't say between my ears

Answer 18

LOL. I can neither confirm nor deny that proposition.
No, you're missing the ds at the end.

Answer 19

haha ok thanks

Answer 20

doing it that way do you get a final change of .8?

Answer 21

wait going back to physics if the distance between the two masses... (s) increase the gravitational force decreases by the same factor right? so if you want the force to decrease by a factor of 0.10 then you would increase the distance by a factor of 0.10.. right?