Simple Physics 12 Problem.

[LiquidTurbo]

I'm looking at a simple Phys12 problem that my cousin asked me to help him with, but I'm surprisingly having some issues with it.

"Earth: m= 5.98e24 kg, moving at speed of 2.97m/s.
Sun: is 1.49e11m away, exerts a force of 3.56e11N on earth.

If the earth orbited the sun in a perfect circle, how much work would the sun do on the earth in one second?"


Who wants to earn some thanks? Thanks in advance.

[!Shuya80]

W = F x D
Fc = mv^2/r

majority of the numbers variables are already there EXCEPT D which you need to find work. Since D is distance. You get the formula D = v*t so in this case in one second you get 2.97m for the variable d.

Fc = (mass of earth * velocity of earth^2)/ distance from the sun.

The result of this equation shall me put in the W=F*D formula.

So we end up with W = Fc * 2.97.

I dont got a calculator handy but it should be right.

[LiquidTurbo]

^

I originally thought that too, but doesn't the F have to be parallel to the motion ? In this case, it's perpendicular, since it's orbiting.

[tiger_handheld]

im glad i didn't take physics !

[kaitamasaki]

I think the solution above is correct
F = mv^2/r is the force exerted on Earth due to the Sun's gravity
and W = F d would give the work done on the Earth by the Sun for that distance, ie. one second.

It is a simple question, maybe just that the concept of gravity performing work is kinda confusing. Its not related to torque or anything so perpendicular or not doesnt matter. I hope I'm not wrong

[ronald55555]

it's 0. The force and velocity vector are always perpendicular.

[!Shuya80]

bah my other post never posted...i was about to say to try plugging it into the whole work formula which is w= FDcos(theta). I forgot to add that sry =.=. I don't quite remember if it's perpendicular that there is no work exerted or no parallell. sorry dude.

[rs_bmw87]

yo it basic too bad you too dumb to do school you wast yor mom money