- #1

AN630078

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- 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

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