Alright, guys. So, in this problem, we're asked to calculate what the period of Mars' rotation is, assuming that we have a satellite in synchronous orbit. Let's go ahead and draw out a diagram quickly. We've got Mars right here, and it has a synchronous orbit around it. I'm told there is a satellite here at some distance, and that satellite is in synchronous orbit with a velocity equal to 1450. What I'm supposed to do is figure out what the period of Mars' rotation is. So, I need to figure out what \( t_{\text{Mars}} \) is. That's really our target variable for this problem. What is \( t_{\text{Mars}} \)? Okay. So, we've got a synchronous orbit. Remember that the period of the satellite orbiting is equal to the period of the planet's rotation. That's what we need to use for this problem. \( t_{\text{Mars}} \) is the variable we're looking for based on the synchronous orbit's equation. Let's start there.
We've got our synchronous orbit equation:
tMars = 4 π r3 G MConsidering all variables, \( G \) (gravitational constant), and \( M \) (mass of Mars) are constants, we are only missing the orbital distance \( r \). We need additional equations to solve for these. Looking at the satellite velocity equation:
V = G M rLet's solve for \( r \):
r3 = G M r2We can rearrange this equation to find \( r \), and using the values provided, we can calculate \( r \):
r = G M V2The value calculated for \( r \) is:
2.04 × 107 meters.Plugging \( r \) back into the equation for \( t_{\text{Mars}} \), then solve and take the square root to find the period of Mars rotation:
t = 7.83 × 109This gives us a rotation period of Mars equal to about 88,470 seconds, or approximately 24.6 hours, which is indeed the correct answer. It's pretty amazing how using two pieces of information, a synchronous orbit and the velocity of that orbit, can help us determine the rotation period of an entire planet. Let me know if you guys have any questions, but that's the final answer. Alright, guys. I'll see you in the next video.