Skip to main content
Ch 13: Newton's Theory of Gravity

Chapter 13, Problem 13

What is the free-fall acceleration at the surface of (a) the moon and (b) Jupiter?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
427
views
1
rank
Was this helpful?

Video transcript

everyone welcome back in this problem, we have the mass of an exoplanet. Okay. We're told that it is triple that of Earth. We're told that it's radius is half of Earth's radius. Okay. And the question wants us to calculate the gravitational acceleration at the exoplanets surface. Okay. Now, when we think gravitational acceleration recall that the gravitational acceleration, Okay, And in this case of the exoplanet, xo is equal to big G. Times the mass of the exoplanet divided by the radius of the exoplanet squared. Alright, Now, filling in some of that information that we've been given. Okay, and again, this G is the gravitational acceleration we're looking for. Okay, so we have big G. The mass of the exoplanet. Well, we're told that the mass of the exoplanet is triple that of Earth. Okay, so let's use that information. So the mass of the exoplanet is going to be three times the mass of Earth. Alright? And then we have the radius, we're told the radius is half of Earth's radius, so one half the radius of Earth. Okay, I'll squirt All right, It's just kind of rearrange here. So we're gonna have three times G. It was a massive Earth. All divided by 1/2 squared is gonna be 1/4. Okay. And then we're left with r squared and that's our of the Earth. Alright, Now we might be looking at this and thinking, okay, I need to find the mass of the Earth. I need to find the radius of the Earth but we can take kind of a more broad approach. Well, we are talking about gravitational acceleration. We know the equation is G. And of the planet over r squared of the planet. In this case we have big G. Massive Earth over R squared of earth. Okay, so this entire quantity here is just going to be a little G of the earth. So what that means is now the gravitational acceleration of the exoplanet is going to be three divided by 1/4. Okay, so that's gonna be 12 times little G of the Earth. It's gonna be 12 times the gravitational acceleration that we experience on earth. Gravitational acceleration on Earth is 9.81 approximately meters per second squared. Okay, and that's gonna give us 117.7 m per second squared. Okay, so the gravitational acceleration of the exoplanet, it's going to be 117.7 m per second squared. That's going to correspond with answer. D Alright, everyone. Thanks for watching. See you in the next video.