Why is the gradient constant in Figure 2?

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Homework Help Overview

The discussion revolves around understanding the gradient of gravitational potential in two figures, particularly why it remains constant in one scenario. The subject area includes gravitational potential, field strength, and the interpretation of distance in gravitational contexts.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between gravitational potential and field strength, questioning the applicability of certain equations in different contexts. There is discussion about the interpretation of distance as radial versus height and how this affects the understanding of the gradient.

Discussion Status

Participants are actively engaging with the concepts, clarifying misunderstandings about the equations involved and the interpretations of distance. Some guidance has been offered regarding the integration of equations and the relevance of context in understanding the figures.

Contextual Notes

There is a noted distinction between the use of radial distance and height, with participants discussing how this affects the interpretation of gravitational potential over small versus large distances. The conversation reflects an exploration of assumptions regarding the nature of the problem presented in the figures.

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nokia8650 said:
I know that V= - GM/r ...
This is your problem. While, strictly speaking, this is true outside of a perfect spherical mass distribution, like a planet, the problem is basically telling you not to use this for Figure 2. Do you know how to get the potential from the field strength?

If you zoom into the curve for Figure 1 so that the percent different between r_min and r_max as very small, what does that part of the curve look like?
 
Ah right, so figure two is esentially the case over a small distance. Is this due to the equation "E=-dv/dr"? Why is this not the case for figure one then, for the larger distance?

Thanks a lot!
 
nokia8650 said:
... so figure two is esentially the case over a small distance.
Yes.
nokia8650 said:
Is this due to the equation "E=-dv/dr"?
No.


nokia8650 said:
... "E=-dv/dr"? Why is this not the case for figure one then, for the larger distance?
It IS the case for both figures. It is just that, in Figure 1, r is interpretted as a radial distance, whereas in Figure 2, r is interpretted as a height. You can obtain V from this equation by integration. BTW, I'm assuming that you realize that this is not an electrostatics problem, but that the similarities are such that you can use the same mathematical construction. So, in your notation, E is the force per test mass, and V is the energy per test mass.
 
Thank you very much, yes, sorry, I meant "g" not "E". I am struggling to understand what the difference between radial distance and height, is it just that height is over a very small distance?

Thanks
 
nokia8650 said:
I am struggling to understand what the difference between radial distance and height, is it just that height is over a very small distance?
It is just an (largely irrelevant) interpretation. It is more phenomenological than physical. For example, if you climb to the next floor of a building, do you imagine that your radial distance from the center of the Earth has increased by 4 meters or that your height above the surface of the Earth has increased by 4 meters. These two interpretations are equivalent; however, the height interpretation is usually more convenient when you are talking about such situations. Conversely, it sounds kind of silly (to me) to say that Mars is 200 million km high today. Typically, we speak of radial distance when the change in distance is on the order of, or larger than, the size of the gravitational bodies, and we speak of height when the change in distance is much smaller than the size of the large gravitational body.
 
That makes sense, thank you ever so much, I really appreciate your help!
 

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