As was foretold, we've added advertisements to the forums! If you have questions, or if you encounter any bugs, please visit this thread: https://forums.penny-arcade.com/discussion/240191/forum-advertisement-faq-and-reports-thread/

Golden Leg
Registered User

So here's the deal.

You are an alien on an alien planet orbitting the planet's sun in a circular orbit. You want to find the mass of your sun. You "triangulate" the center-to-center distance between your planet and sun to be 3.52E+10 meters. The period of motion of your planet (the length of your year) is 1.81E+7 seconds. You know G = 6.67E−11 Nm^2/kg^2. What is the mass of your sun?

Given: R = 3.52E+10 m, T = 1.81E+7 s, G = 6.67E−11 Nm^2/kg^2

Want: M (as opposed to m)

Equation: F = GmM/r^2, ?, ?

We have three unknowns, M, m, and F. With T, one could potentially solve for angular velocity (omega = 2PI(1/T)) and go from there.

Anyone up for some phun?

You are an alien on an alien planet orbitting the planet's sun in a circular orbit. You want to find the mass of your sun. You "triangulate" the center-to-center distance between your planet and sun to be 3.52E+10 meters. The period of motion of your planet (the length of your year) is 1.81E+7 seconds. You know G = 6.67E−11 Nm^2/kg^2. What is the mass of your sun?

Given: R = 3.52E+10 m, T = 1.81E+7 s, G = 6.67E−11 Nm^2/kg^2

Want: M (as opposed to m)

Equation: F = GmM/r^2, ?, ?

We have three unknowns, M, m, and F. With T, one could potentially solve for angular velocity (omega = 2PI(1/T)) and go from there.

Anyone up for some phun?

0

## Posts

this sounds more like

someone'shomework than phunthisisntwallyonGolden Legonthisisntwallyonyou are given a circular orbit which is pretty good; that makes things easy. I don't think Newton's law of universal gravitation is the right way to go on this one. this was just a quick google search so ill just hook you up here. Apparently:

the orbital period of something in a circular orbit is: T=(2pi)(r^3/(G*M))^(1/2). where M's the mass of the thing you are orbiting. You should be able to algebrate that and solve for what you need.

FuzzywhaleonSir Headless VIIonAbsolutely nothing.

SentryonshadydentistonGT: Tanky the Tank

Black: 1377 6749 7425

The equation F = GmM/r^2 can be set to solve for M, rendering M = Fr^2/Gm. F is equal to the centripetal force, and Fcentripetal = (OMEGA)^2(r). Angular velocity, or OMEGA, equals 2PI/T. Solving for OMEGA and plugging into the equation for M gives us M = [(OMEGA)^2*(r)^3]/G.

Turns out the mass of that sun equals 6.55x10^29 kg.

Golden LegonAridholonlocomotivemanonI actually worked at work on Saturday. Also I went out on a date with a real life girl.

Can you like, permanently break the forums?