Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.50 s. The period on Mars turns out to be 2.45 s. What is the free-fall acceleration on Mars?

Video transcript

Welcome back, everyone. We are making observations about a self navigating space vehicle that contains a pendulum. Now, this vehicle moves between planetary objects and it starts out on Jupiter where the period of the pendulum on Jupiter is 1.2 seconds. And Jupiter we are told has a gravitational acceleration of 24.8 m per second squared. Now, the vehicle then moves over to a different unknown planetary object where we are told the pendulum has a period of 3.2 seconds and we are tasked with finding what is the gravitational acceleration on that planet. Well, we have a formula that for the period of a pendulum and we're going to apply it to both scenarios here. So on Jupiter, we have that the period of Jupiter is going to be equal to two pi times the square root of the length of the cable involved in the pendulum divided by the gravitational acceleration of Jupiter. Similarly, for this unknown planetary object, it's the same formula except we're just plugging in the planetary values for our a period as well as our gravitational acceleration. So this will be two pi times the square roots of L divided by GP, which is our desired value. Now to isolate GP, what I'm gonna do is I'm actually gonna divide our period on Jupiter divided by our period on our unknown planetary object. And we get that TJ divided by T P is equal to two pi times the square root of L all divided by the square root of B J. All I did was uh separate the radical through the numerator and denominator, which you definitely can do divided by two pi times the square root of L divided by the square root of GP. And when you divide by a fraction, what you really do is multiply by its reciprocal. And so what I'm gonna do is I'm actually going to move this up to be our numerator and then I'm gonna move it, the former numerator, two pi square root of L to be our new denominator. We'll get rid of those lines redraw. And this is what we are working with here. Two pi times a square root of L cancels out on top and bottom. And what we are left with is the square root of GP divided by the square root of G J. We want to get rid of these radicals here and we want to isolate our gravitational acceleration on our unknown planetary object. So what I'm gonna start off by doing is squaring both sides of our equation. And what this does is this gets rid of the radicals on the right hand side of our equation. And finally, to get rid of our gravitational acceleration on Jupiter, I would just multiply both sides by G J. Our G J S cancel out on the right hand side. And what we are left with is that GP is equal to G J times TJ. Oops, sorry, that should be TJ here, TJ divided by T P all squared and we know all these values. So let's go ahead and put in our values here. We have G J is 24.8 times 1.2 divided by 3.2 squared. And when you plug this in our calculator, we get 3.49 m per second squared for the gravitational acceleration on our unknown planetary object, which corresponds to our final answer. Choice of B Thank you all so much for watching. I hope this video helped see you all in the next one.