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I'm combining three equations from my physics text book:

  • Newton's law of gravitation: $F = -\frac{GMm}{r^2}$

  • The centripetal force equation: $F = \frac{mv^2}{r}$

  • The equation for the speed of an object traveling in a circle: $v = \frac{2 \pi r}{T}$

I wanted to create an equation to find the Time period, $T$ and ended up with: $T = \frac{2 \pi r^2}{GM}$ Which is wrong...

EDIT

I've worked it out again, this is my working:

I put Newton's law of gravitation and the centripetal force equation equal to each other:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Multiply both sides by $r$:

$\frac{GMm}{r} = mv^2$

Sub in $v = \frac{2 \pi r}{T}$ for $v$:

$\frac{GMm}{r} = m(\frac{2 \pi r}{T})^2$

Divide both sides by $m$:

$\frac{GM}{r} = (\frac{2 \pi r}{T})^2$

Root both sides:

$\sqrt{\frac{GM}{r}} = \frac{2 \pi r}{T}$

Flip both sides and divide by $2 \pi r$:

$T = \frac{2 \pi r}{\sqrt{\frac{GM}{r}}}$

EDIT 2 Which I can simplify:

Multiply both sides by $\sqrt{\frac{GM}{r}}$:

$T \times \sqrt{\frac{GM}{r}} = 2 \pi r$

Square both sides:

$T^2 \times \frac{GM}{r} = (2 \pi r)^2$

Divide both sides by $\frac{GM}{r}$:

$T^2 = \frac{(2 \pi r)^2}{\frac{GM}{r}}$

Clean it up:

$T^2 = \frac{(2 \pi r)^2 \times r}{GM}$

Take out $r$ to get the final answer:

$T^2 = \frac{(2 \pi)^2}{GM}r^3$

If you take out the constant you get Kepler's law (as Ross Millikan said):

$T^2 \propto r^3$

Is this correct? It's going in my A-Level Physics notes and I don't want to be learning the wrong stuff when it comes to the exam. This is correct now, thanks guys!

If anybody's interested, I've open sourced the notes here

Prajnan Bhuyan
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Todd Davies
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3 Answers3

5

It is not correct. Kepler's third lawstates: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Your solution has the square, not the $\frac 32$ power of the axis. You are using the correct input, so if you show your work we may find the problem. $v$ should be proportional to $\frac 1{\sqrt r}$

Ross Millikan
  • 374,822
2

Since centripetal force exert on the orbited body is equal to the gravitational force exert between the two bodies, then

$$F_c=F$$ $$mv^2/r=GMm/r^2$$ $$v^2/r=GM/r^2$$ [since m and m do cancel] $$w^2r=GM/r^2$$ (since v=wr) $$w^2=GM/r^3$$ $$4p^2/T^2=GM/r^3$$ [where is a radian pie] $$T^2=r^34p^/GM$$

Therefore period $T=\sqrt{4p^2*r^3/GM}$.

peterh
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Christian Nding'wa
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F(g)=GMm/(R+H)^2 (gravitational force) F(centripetal force) =Mv^2/R+H (here v is orbital velocity) here centripetal force is provided by gravitational force therefore GMm/(R+H)^2=Mv^2/R+H by working out we get v^2=Gm/R+H H is comparatively small than R (radius of earth) and is negligible therefore v= squareroot of GM/R

but g=GM/R^2 (g is acceleration due to gravity) or GM/R=Rg

atlast v =squareroot of Rg(where v is orbital velocity)

Sudeep Gumaste
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