Gravity problem with non constant acceleration

Click For Summary

Homework Help Overview

The problem involves a scenario where a rock is dropped into a tunnel drilled through the center of a uniform solid sphere, representing a planet. The focus is on understanding the changing acceleration due to gravity as the rock moves towards the center and then to the opposite side of the planet.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to derive the mass of the planet and question the relevance of escape velocity in this context. There is exploration of the gravitational force acting on the rock and how it changes as the rock descends through the tunnel.

Discussion Status

The discussion is ongoing, with participants providing insights into the gravitational forces involved and questioning the assumptions made about the mass of the planet and the variables at play. Some guidance has been offered regarding the gravitational attraction formula and its dependence on the mass of the sphere at varying distances.

Contextual Notes

There is a noted confusion regarding the application of gravitational equations, particularly about which mass should be considered when the rock is at different distances from the center of the planet. Participants are also addressing the implications of the problem's setup on the calculations needed.

inferno298
Messages
25
Reaction score
0

Homework Statement



You are a space traveler and you land on the surface of a new planet. You decide to drill a tunnel through the center of the planet to the opposite side of the planet. Your idea is to use this tunnel to easily transport rocks from your location to the opposite side of the planet. So, you drop a rock in the tunnel and it falls down the hole toward the center of the planet, passes through the center of the planet, then makes it to the opposite side of the planet. Assume the planet is a uniform solid sphere of radius 8.50×10^6 m. The rock is dropped from rest into the hole, and it initially has an acceleration of 10.70 m/s2. How long will it take to make it to the other side of the planet?

Homework Equations



F = (G m1 m2)/r^2
U = -(G m1 m2)/r

The Attempt at a Solution



So I guess I am having a hard time starting this, which in essence is pretty much the only hard part of any physics problem. I pretty sure this will be a differential equation since the acceleration due to gravity will be changing as the object gets closer to the center of the sphere. I also believe that this might be able to be solved as a simple harmonic oscillator diff'eq.

First ill try and get the mass of the planet. Using a and r I can solve for the escape velocity and then use that to solve for the mass. Using equations:
Ve = Sqrt[2*a*r]
M = (Ve^2 * R)/(2*G)

Now F = m1 a
m1 = mass of stone
m1*a = (G m1 m2)/r^2

m1 cancels on both sides
a = dv/dt
dv/dt = (G m2)/r^2

so a will change but so will r, that's two changing variables since r is a function of t also. Would I have to do some Lagrange Equations of Motions? This where I am lost at. Don't know if I am making this more complicated then it needs to be.
 
Physics news on Phys.org
Where did that escape velocity came from? I don't see it mentioned in the problem.
You could find the mass of the planet by using the value of the surface acceleration of gravity but the answer does not depend on the mass so finding the mass is not really useful.

When you write the force of gravity on the m1, what is the significance of m2?
 
it was just a round about way to get the mass of the planet, escape velocity didn't need to be mentioned. All the tools were there. I just forgot the direct way to get to the mass. As for m2, that is the mass of the planet, m1 cancels out. m2 didn't cancel out, therefore it was still part of the Gravitational central force equation.

Any more help on how to get going would be appreciated
 
m2 should not be the entire mass of the planet.
When the body is at distance r from the center (in that tunnel), with r<R, only the mass of the sphere of radius r enters the formula for gravitational attraction. Here R is the radius of the planet.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
23
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K