Apollo astronauts took a 'nine iron' to the Moon and hit a gol...

Question

Apollo astronauts took a 'nine iron' to the Moon and hit a golf ball about 180 m. Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit it only 32 m, estimate the acceleration due to gravity on the surface of the Moon. (We neglect air resistance in both cases, but on the Moon there is none.)

Accepted Answer

Welcome back. Everyone in this problem. A space explorer conducts tests on a remote asteroid, employing a tennis racket to strike a tennis ball. The ball traverse is roughly 120 m across the asteroid's terrain given that the swing launch angle and other variables remain consistent if the same explorer were to strike the ball on earth where it typically travels only 25 m deduce the acceleration due to gravity on the asteroid's surface. Please note that air resistance is insignificant on both the asteroid and earth and assume that the tennis ball takes off and lands at the same height for our answer choices. A says it's 9.8 m per second squared. B 4.9 m per second squared, C 3 m per second squared and D 2 m per second squared. Now we're trying to deduce the acceleration due to gravity for our asteroid. So first of all, let's make sense of this problem with a diagram. So we have two situations here. We're talking about the space explorer on an asteroid. OK. So let's say that this is our asteroid, the surface of our asteroid here compared to the space explorer conducting the test on earth. OK. So we're throwing a ball on both of these surfaces. Now, we're saying that on the earth, if we, when whenever he strikes the ball using a tennis racket, it travels 25 m, talking about its horizontal distance here. So we strike the ball and it does 25 m, ok. Now our ball is right here. OK. So that distance is 25 m. But in comparison on the asteroid, the ball travers is 120 m after he strikes the ball. So we can imagine that its horizontal distance probably looks something like this, of course, not drawn to scale, but just to give us an idea of the comparison here. OK. And we know that it is our our our acceleration due to gravity that makes the difference. So the question is how can we figure out what the acceleration due to gravity is on the asteroid? Knowing that the acceleration due to gravity on the earth, let's call it ge is equal to 9.8 m per second squared. So that's basically what we are trying to find. Now, let's think about it here. We have information about the range of the ball. It's horizontal distance. What do we know about range and gravity? Well, recall, OK, recall that for a projectile in motion, its range is equal to V not squared multiplied by the sine of two theta. OK. Divided by the acceleration due to gravity where R is the horizontal range, theta is our launch angle and G is our acceleration due to gravity like we had said. And V not is our initial velocity. I can't remember if I just said that, but I think I did now, if we're thinking about earth, OK. If we're thinking about the range for earth, then the range, let me put that in blue here. Then the range for earth, let's call it re is going to be equal to initial velocity V knot squared multiplied by the sine of two theater. OK? Divided by the gravity on earth. OK? And here we're assuming that it's still in the same, it's the same initial velocity, all the variables are consistent. So it's the same launch anger the same swing and so on. Likewise for the range on the asteroid, OK? The range on the asteroid is going to be equal to V knot squared sine two theta. OK? And that's going to be divided by the gravity on the asteroid. OK. So now we know what the range is. We have a formula that connects the gravity to the range in both situations. But here's the thing when we look at both of these equations, we can compare them even further for the acceleration due to gravity because by dividing both formula OK, let's say we were to divide both formula OK? And dividing them because now we're getting a ratio of the range on earth compared to the range on the asteroid, it's going to be equal to V not squared multiplied by sine two theta. OK. Divided by gravity on earth, divided by the range on the asteroid, which means we're multiplying by reciprocal. So it's going to be multiplied by G divided by vno squared sine to theta. OK. Now notice here that the term in the numerator of the range on the earth cancels the denominator for the range on the asteroid. So what this tells us then is that the range on the earth, the rings on the earth? OK. The range on the earth to the range on the asteroid is going to be equal to the gravity on the asteroid to the gravity on the earth. So that's the ratio that we can come up with. Ok. So the ratio of the range on the earth to the asteroid is the same as the gravity of the asteroid to the earth. Now think about it. We have the range of the earth, the range of the asteroid, the the acceleration due to gravity on the earth. We just need to find the gravitation due to the acceleration due to gravity on the asteroid. So we can substitute the values to solve because no, that means the gravity on the asteroid divided by the gravity on the earth which is 9.8 m per second squared is equal to the range on the earth. 25 m divided by the range on the asteroid 120 m. That's the case, that means the gravity on the acceleration due to gravity on the asteroid is going to be equal to 9.8 m per second squared multiplied by 25 m divided by 120 m. No, when we work that out, when we put it into our calculator, we'll find that the acceleration due to gravity on the asteroid is equal to 2 m per second squared. If we look back on our answer choices, that means the correct answer is answer choice D. And when we think about the solution, it really does make sense. If the asteroid is able to travel fur or the ball is able to travel further on the asteroid, that means its gravity must be weaker than the gravity on earth. So that's one way we can know that we are correct. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Acceleration due to Gravity

Projectile Motion

Kinematic Equations

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