I If a rock fell 20,000 meters to Earth Question

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A rock falling 20,000 meters to Earth, assuming no air resistance and only Earth's gravitational influence, raises questions about the time it takes to hit the ground due to changing acceleration. The gravitational force can be described by the equation g=m1(m2)G/r^2, which complicates the calculation as the rock approaches Earth. While some participants suggest using elliptic integrals for a precise solution, others argue that the difference in time calculations using constant versus variable gravity is minimal, only about 0.17 seconds. The discussion emphasizes the importance of setting up a free body diagram and integrating the equations of motion to find the time of fall. Overall, the problem illustrates the complexities of gravitational physics and the nuances of mathematical modeling in real-world scenarios.
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I was thinking about this question:

So if a rock fell 20,000 meters to Earth assuming no air resistance and no other gravitational forces from anything but Earth, how long would it take the rock to hit Earth? Keep in mind acceleration is changing through the gravity equation as the rock gets closer and closer to Earth -> g=m1(m2)G/r^2.

(I'm not sure if its necessary but if it is, say the rock is 5 kg).

I put undergrad because this was not a question that came up in class and I think it involves integration.

This is not a hw/cw question it is a question I came up with on my own.
 
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Welcome to PF, Jack. Do you have any specific questions about the question you were thinking about?
 
Hello, I wanted to know if anyone had a possible solution to the question which is how much time it would take to hit Earth. I've been trying to figure it out but can't figure out how to fit time into the overall equation because I always just get the equations in terms of radius.
 
If you search PF, sometimes you can find similar inquiries. @HallsofIvy did a good development on this topic in this PF thread. I must admit I stall out on the subject of the elliptic integral. Maybe you can take it further.
 
Am I missing something here? Why do you need an elliptic integral? Set up a free body diagram where the force found from Newton's law of gravity is equal to the mass of the ball times its acceleration. The equation is separable and so is fairly easily integrated (twice) to get the separation between the two bodies as a function of time. Maybe I am missing something here?
 
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Jack981 said:
So if a rock fell 20,000 meters to Earth assuming no air resistance and no other gravitational forces from anything but Earth, how long would it take the rock to hit Earth? Keep in mind acceleration is changing through the gravity equation as the rock gets closer and closer to Earth -> g=m1(m2)G/r^2.
20 km? How much does g really change? Do you really need to bother with that? If you decide no, do you know how to do the problem then?
 
russ_watters said:
20 km? How much does g really change? Do you really need to bother with that? If you decide no, do you know how to do the problem then?

The time difference between calculations using constant surface and variable gravity for ##20km## is only ##0.17s##, I believe.
 
Jack981 said:
I was thinking about this question:

So if a rock fell 20,000 meters to Earth assuming no air resistance and no other gravitational forces from anything but Earth, how long would it take the rock to hit Earth? Keep in mind acceleration is changing through the gravity equation as the rock gets closer and closer to Earth -> g=m1(m2)G/r^2.

(I'm not sure if its necessary but if it is, say the rock is 5 kg).

I put undergrad because this was not a question that came up in class and I think it involves integration.

This is not a hw/cw question it is a question I came up with on my own.

https://www.physicsforums.com/threa...ith-varying-acceleration.866218/#post-5437928
 
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