Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93 * 10^22 kg and a radius of 1821 km. For this calculation, ignore any variation in gravity over the 500-km range of the debris. How high would this material go on earth if it were ejected with the same speed as on Io?

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Welcome back everybody. We are looking at projectile experiments both on the surface of the Earth and on the surface of MARS. Now we are told a couple of things. So we're going to launch a projectile into the atmosphere on Earth. We're going to do the same on MARS but we're actually given information about the projectile on Earth. We are told that its total height achieved is kilometers or 10,000 m right now, since we're on Earth, the acceleration that we're dealing with is just going to be the acceleration due to gravity, which is equal to negative 9.8 m per second squared. Now, at the top of projectile motion, we know that our velocity is just going to be Z zero. So if we're looking at the frame of reference, starting down here, going up to the top, our final velocity will be equal to zero. Now we are told that we are going to replicate this launch on Mars and we need to figure out how high the projectile is going to launch on Mars with the same initial velocity but what is the initial velocity? Well, we're gonna have a couple of steps to this problem here and let me write it out, Step one, we're going to need to find what our initial velocity is for both launches. We're also going to then have to find what the gravitational acceleration on MARS is since we are told that MArs has a mass of 6.42 times 10 to the 23rd kilograms in a radius of kilometers. We have to figure out what that gravitational acceleration is and then finally, we will be able to find our delta y or how high the projectile travels on MArs. Now the key to this entire equation, we're dealing with velocities, we're dealing with accelerations is going to be our kid a Matic equations. And the one that comes to mind for me is this one that our final velocity squared minus our initial velocity squared is equal to two times our acceleration times the height that the projectile travels. We're gonna use this equation multiple times throughout this problem. But let's go ahead and get started with part a using this equation right here, I'm actually gonna rearrange it a little bit to actually isolate our initial velocity right here. Right, So let's see, what do we have, what do we have? We are going to have that our initial velocity is going to be equal to the square root of negative two times our acceleration times our delta y. Now we don't know anything on mars about the projectile but we have all of these values on Earth and keep in mind we're going to be using the same initial launch speed. So let's Just go ahead and plug in the values that we know right? So this is equal to the square root of negative two times our acceleration due to gravity is negative 9. times the height that is covered, which is 1010 km or 10,000 m. We need to work in meters for this. This gives us an initial velocity of 442.7 m/s. Great. So now we have our initial velocity but we still need this acceleration in order to have all of our values to calculate our delta Y on MArs. Well, that's where this comes in. So let me switch colors here for part B R. Acceleration that we want is our gravitational ax acceleration on MArs. Now this is given by this formula of Big G times the mass of MArs divided by the radius. Big G here is just some constant equivalent to 6.673 times 10 to the negative 11th Times our mass on Mars which is 6.4, 2 times 10 to the 23rd all over our radius which we are told is 3390 kilometers. But we need to work in meters. So we're gonna multiply that by 1000. When you plug this into your calculator, you get that are gravitational acceleration on MArs is 3.72. Sorry, let me write that a little bit neater. 3.72 m per second squared great. So now we have our initial velocity. We have our acceleration, our final velocity is going to remain the same as our projectile was on earth. It's still going to be zero. So now we can go ahead and isolate delta Y in part C. Here. So, rearranging this equation again, this time we get that our delta Y on our MARS is going to be equal to our final velocity squared minus our initial velocity squared all divided by two times our acceleration. This is equal to zero squared minus our velocity that we found of 442.7 squared, Divided by two times our acceleration, Which it's going to be negative because remember, it does point downwards, even on Mars, it points towards Mars. So this gives us an answer of 26,300 m or 26.3 km, giving us a final answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

Verified Solution

Key Concepts

Gravitational Acceleration

Kinematic Equations

Conservation of Energy