"A satellite orbits the sun with a period of 1.0 day. What is the radius of its orbit?"

Question

Accepted Answer

Hey, everyone. So this problem is dealing with gravitational force. Let's see what it's asking us. An artificial satellite with mass M is in orbit around a distant planet Rubiks of mass 4.5 times 10 to the 25 kg. The asteroid's orbital period around planet Rubiks is two days determine the radius of its orbit. Our multiple choice answers here are a 3.81 times 10 to the six m B 1.82 times 10 to the eight m C 3.34 times 10 to the six m or D 1.31 times 10 to the eight m. OK. So the first thing we can do here is recall that our gravitational force equation is given by F equals G M one M two over R squared where G is our gravitational constant M one and M two are the masses of the two objects. So in this case, it'll be our asteroid and planet Rubik as the two masses. And then R is the distance between those two masses. It's important here to recognize that the distance between those two masses is actually the radius of the asteroids orbit around the planet. So that R is what we are solving from Newton's second law, we can recall that force is equal to mass multiplied by acceleration. And then we're going to recall our uniform circular motion equations. One for centrical acceleration and then one for velocity. So our acceleration is given five V squared over R in our velocity V is given by two pi R over T. So when we combine that we get an acceleration uh four pi squared R over T sorry, the over T square. So we're going to substitute mass multiplied by acceleration and then plug in this acceleration using our uniform circular motion equations into this first equation. So we have G M one M two over R squared is equal to M. Now this is the mass of the asteroid. So that will be M one multiplied by four by squared are all divided by two squared. So the masses of the asteroid here cancel. And then we can simplify our equation to solve four R. So we'll have G multiplied by M two multiplied by T squared divided by four pi squared. And that equals R qed. So from here, we can just double check that we have all of the values that we need to solve this problem. So G is our gravitational constant. We can recall that that is 6.67 times 10 to the negative 11 meters cubed per kilogram second squared, the mass of our planet was given to us in the problem as 4.5 times 10 to the 25 kg. And tea is our orbital time and that was given as two days. Now, we need to get that into standard units of seconds. So we will multiply that by 3600 seconds per hour, 24 hours in a day. And then you can convert that to seconds, plug that into our calculator and we get 1.73 times 10 to the fifth seconds. So now we do have everything we need to solve for our. So R is going to be the cubed root um 6. times 10 to the negative 11. And that's meters cubed per kilogram second squared multiplied by our mass 4.5 times 10 to the 25th kilograms multiplied by our time, better run out of space here. So forgive me, time is gonna be 1.73 times 10 to the fifth seconds. And that's gonna be squared and then all over four pi square. So when we take the cubed root of that, we get 1.31 times 10 to the eight m. And so that is the answer to our problem. And when we look at our multiple choice answers that aligns with answer choice D. So that's all we have for this one. We'll see you in the next video.

"A satellite orbits the sun with a period of 1.0 day. What is the radius of its orbit?"

Verified Solution

Key Concepts

Kepler's Third Law

Gravitational Force

Circular Motion