Rearranging gravitational and centripetal equations to derive formulas

[AN630078]

Homework Statement

Hello, I have been tasked with rearranging various equations to derive certain formulas but I do not know whether I have done so correctly, or necessarily using the correct quantities. Would anyone be able to look over my workings and possibly offer some advice?

1. Using Newton’s second law of motion and a formula for centripetal acceleration, write down a formula for the centripetal force on a planet of mass m moving in a circular orbit of radius r with a speed v.

2. Equate your last answer to the force as shown by Newton’s law of universal gravitation, between a planet of mass m at a distance r from the Sun of mass M. Hence show that as r decreases v increases.

3.By using the other formulae for centripetal acceleration, show that T squared is proportional to r cubed.

Relevant Equations

F=ma
ω=2π/T
a=rω^2 = v^2/r
F=GMm/r^2

1. Newton's Second Law states F=ma and the formula for centripetal acceleration is v^2/r
Therefore, F= mv^2/r

Would this be complete, I just feel that I should need to do something further but I am not sure what?

2.F=mv^2/r
Gravitational force = GMm/r^2
Gravity is the cause of centripetal force so these are equal;
mv^2/r=GMm/r^2
Cancel m; v^2/r=GM/r^2
Multiply by r; v^2=GM/r

v^2∝1/r
or v∝1/√r

GM is constant so as v increases r decreases.

3. I would assume that the question here means a= rω^2
Therefore, F=mrω^2
The angular velocity is ω=2π/T

The gravitational force is given by F=GMm/r^2 and since the planets moves in a circular orbit, consider the centripetal force also, F=mω^2r. As both gravitational and centripetal forces act in the same direction, one can equate them to find GMm/r^2=mω^2r.
Note that 'm' cancels; GM/r^2=ω^2r
Divide by r: GM/r^3=ω^2

Substitute the value of the angular velocity into the equation; GM/r^3=(2π/T)^2 which becomes GM/r^3=4π^2/T^2
Multiply both sides by T^2; T^2 GM/r^3=4π^2
Multiply both sides by r^3; T^2 Gm=4π^2*r^3
Divide both sides by GM; T^2 =4π^2*r^3/GM

Since 4π^2/GM is constant T^2 is indeed proportional to r^3.

Thank you very much to anyone who replies

 

[haruspex]

As both gravitational and centripetal forces act in the same direction, one can equate them

It takes a bit more than that... the gravitational force is supplying the centripetal force, as no other force acts in that direction.

Other than that, all good.

 

[AN630078]

It takes a bit more than that... the gravitational force is supplying the centripetal force, as no other force acts in that direction.

Other than that, all good.

Thank you very much for your reply, do you know I was unsure about that statement before posting so thank you for picking up on that. Do you think I have used the correct equations in the questions specified, e.g. "the other formulae for centripetal acceleration" etc. ? Are there any improvements I could make?

 

[Steve4Physics]

I'd like to add a few words of clarification too.

Are there any improvements I could make?

Good answer overall but it is incorrect to say:
“both gravitational and centripetal forces act in the same direction"
because it suggests there are 2 different forces acting. There is only one. The centripetal force *IS* the gravitational force in this question.

It is a common error to think that centripetal force is a special type of force, but it is just the name we give to whatever force happens to be acting perpendicular to the velocity. Such a force (be it gravitational, electrical, magnetic or the resultant of several other forces) causes the direction of velocity ot change and the circular path is the result - with the 'centripetal' force pointing towards the centre of the circle.