Gravity(Part 2)

Simple Circular Orbit

Consider an object with mass m orbiting a planet with mass M, at a distance r from the planet, in a circular orbit, with velocity v.  

 

 

 

You  may see a VPython animation of a circular orbit here:

https://www.glowscript.org/#/user/zorniy/folder/Public/program/CircularOrbit 

 

Or a little video of the program running:

 

 

 

There are two forces acting on the object:

• Gravity, pulling inwards towards the planet
• Centrifugal force, acting outward

In a circular orbit, the distance r remains the same, meaning there is no acceleration outwards or inwards.

According to Newton's Second Law:

$F_{total} = ma$

gravity + centrifugal = 0

$-\frac{GMm}{r^2} + \frac{mv^2}{r} =0$

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

$v^2 = \frac{GM}{r}$

We obtain the velocity of the object orbiting in a circular orbit:  

$v = \sqrt{\frac{GM}{r}}$

This velocity must be tangential to the orbit, or at right angles to the vector r between the object and the planet.

Orbital Period

The orbital period T is the time taken to complete one full orbit. In physics, the result of the calculation will be in seconds.

 

If we multiply the orbital velocity v with the orbital period T, we get the total distance traveled, which is equal to the circumference of a circle.

$vT = 2\pi r$

 

$T = \frac{2 \pi r }{v}$

 

But  $v = \sqrt{\frac{GM}{r}}$

 

$T = 2 \pi r \sqrt{\frac{r}{GM}} $ 

So we arrive at the orbital period:

$T = \frac{2 \pi}{\sqrt{GM}} r^{3/2}$

We can calculate the orbital period T if we know the mass of the central mass M and the distance r.

 

But we can also use it to calculate the mass of a planet, if we can measure the orbital period of its satellite and the orbital distance. 

This is how the masses of planets and other stars were calculated; astronomers observed something that orbited them. Astronomers measured the time for the satellite to complete one orbit, and measured the separation between the satellite and the planet.

$T^2 = \frac{4 \pi^2}{ GM }r^3 $

$M = \frac{4\pi^2}{GT^2}r^3$

 

There is a moon called Ganymede orbiting the planet Jupiter. Its orbital period is 172 hours, and it orbits 1,070,400 km from Jupiter.

$T = 172 \times 60 \times 60 s = 6.19 \times 10^{5} s$

$r = 1.07 \times 10^9 m$

$G = 6.67 \times 10^{-11}$

Mass of planet Jupiter:

$M = \frac{4\pi^2}{6.67 \times 10^{-11} \times (6.19\times 10^5)^2}\times (1.07\times 10^9)^3 = 1.89\times10^{27}kg$

 

 

This is in good agreement with the value in Wikipedia: $1.898 \times 10^{27} kg$

Do the following:

 

 

Google distance of the Moon from Earth, orbital period T of the Moon around Earth. 

• Convert to meters and seconds.
  
• Calculate the mass of Earth using the equation above. 
• Compare with the mass of Earth according to Google. 

Google distance of Earth to the Sun, and the orbital period.

• Convert to meters and seconds.
  
• Calculate the mass of the Sun using the equation above. 
• Compare with the mass of Sun according to Google.