Setting up integral for a one-dimensional planet

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Homework Help Overview

The problem involves determining the gravitational field at a point located along the x-axis due to a uniform rod of mass M and length L. The rod is centered at the origin, and the point of interest is at a distance d from the origin, where d is greater than half the length of the rod.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up the integral for the gravitational field by expressing the gravitational contribution from a small mass segment of the rod. They question the correct expression for the distance r from the mass segment to the point where the field is being calculated.

Discussion Status

Some participants provide guidance on visualizing the problem by suggesting the drawing of a diagram and clarifying the definition of r. The original poster appears to have made progress in understanding the limits of integration for the problem.

Contextual Notes

The discussion includes considerations of uniform density and the relationship between mass segments and the total mass of the rod. There is an implicit assumption that the gravitational field can be calculated using integration techniques.

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Homework Statement


A uniform rod of mass M and length L lies along the x-axis with its center at the origin. Determine the gravitational field at the point x=d where d>L/2.


Homework Equations





The Attempt at a Solution


I know I will have to take the integral of dg in order to find g. g=Gdm/r2.

It has uniform density, so λ=M/L

So, dm=(M/L)dr

Now, the problem is finding r.

I called the distance from the origin to the point: d, and the distance from the end of the rod to the point: x. Is r=d? And then you integrate from (-L/2) to (L/2)?

Thank you.
 
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Draw a picture of your rod. Now mark off the little mass segment, dm.

r is the distance from dm to the point at which you want to calculate the field.
 
I did that, and I've figured it out. r is r, and you integrate from d-L/2 to d+L/2
 
musicmar said:
I did that, and I've figured it out. r is r, and you integrate from d-L/2 to d+L/2

Cool! :smile:
 

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