A recently discovered extrasolar planet appears to be rockier and...

Question

A recently discovered extrasolar planet appears to be rockier and denser than earth. It is 16 times as massive as earth, but its diameter is only twice that of earth. What is the free-fall acceleration on the surface of this planet?

Accepted Answer

Welcome back everybody. We are told that the radius of some exoplanet is 0.9 times the radius of Earth. Were also told that the mass of the exoplanet is 0.0 to three times the mass of Earth. And we are asked to find what the gravitational Acceleration is of that exoplanet. Well, the easiest way to do this is going to be as we did above. Right to find some constant ratio between the two times the gravitational acceleration of Earth, which we know to be 9.81. Well, the gravitational acceleration of a given planet is equal to Newton's gravitational constant times the mass of the planet, all divided by r squared. So here's what we're gonna do to find that ratio or to find that constant C. We're gonna take the formula for the gravitational acceleration of the exoplanet and divide it by the gravitational acceleration of Earth. Right now we know what the gravitational acceleration of Earth is, but we're still going to use this formula for both because we'll be able to use the given values above and you'll see why. So when you divide the fractions and you know when you divide by a fraction, you flip it multiplied by its reciprocal and whatnot. You're going to get this that are constant is equal to the mass of the extra planet divided by the mass of the earth times the radius of the earth divided by the radius of the exoplanet squared. Now, what are these ratios? We'll just given by these constants. Now, be careful here. As you see, the radius of Earth is on top, meaning that it's going to be the reciprocal of this value right here. So let's plug in our values. We have that the ratio of the masses is 0. to three times the reciprocal of our um ratio between the raid E. I. So this will be 1/0 0.9 squared. Giving us that R. C. Or constant value is 0.28. Now that we have that, let's go ahead and find our gravitational acceleration of the exoplanet is going to be 0.28 times the acceleration due to gravity on Earth, which we just know to be 9.81 is equal to 0.27 m per second corresponding to answer choice C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

A recently discovered extrasolar planet appears to be rockier and denser than earth. It is 16 times as massive as earth, but its diameter is only twice that of earth. What is the free-fall acceleration on the surface of this planet?

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