Work done on a meteor approaching the Earth -- Is it +ve or -ve?

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Discussion Overview

The discussion revolves around the question of whether the work done by gravity on a meteor approaching Earth is positive or negative. Participants explore the mathematical formulation of work in the context of gravitational forces and the implications of the sign of work on the meteor's velocity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation of work done by gravity using the integral of the gravitational force, suggesting that the result is positive based on their interpretation of the formula.
  • Another participant emphasizes the need to consider the dot product of force and displacement vectors to determine the sign of work, indicating that the work can be negative if the vectors are in opposite directions.
  • Further clarification is provided regarding the relationship between the direction of motion and the change in position vector, with a focus on the angle between the vectors affecting the sign of work.
  • Some participants agree that if the work done by gravity is negative, it implies that the velocity of the meteor should decrease.

Areas of Agreement / Disagreement

Participants express differing views on the sign of the work done by gravity, with some suggesting it is negative while others initially propose it could be positive. The discussion remains unresolved regarding a definitive conclusion on the sign of the work.

Contextual Notes

There are nuances regarding the definitions of direction of motion and position vectors that are not fully resolved, and the implications of the sign of work on the meteor's velocity are also discussed but not conclusively agreed upon.

Ben Harris
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Hi all, i am trying to figure out if work done by gravity is positive or negative on an approaching meteor,

I have used W(1->2) = integral (1->2) F(gravity) dot dr. and got a result of (GMem/r2) - (GMem/r1).

Were G is gravity, Me is Earth mass, m is meteor mass and r1, r2 are initial and final radius respectively.

As (GMem/r2) > (GMem/r1) the result has to be positive right?

any help is appreciated :)

thank you
 
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Ben Harris said:
Hi all, i am trying to figure out if work done by gravity is positive or negative on an approaching meteor,

I have used W(1->2) = integral (1->2) F(gravity) dot dr. and got a result of (GMem/r2) - (GMem/r1).

Were G is gravity, Me is Earth mass, m is meteor mass and r1, r2 are initial and final radius respectively.

As (GMem/r2) > (GMem/r1) the result has to be positive right?

any help is appreciated :)

thank you
It's simpler than that. The work done by a force on an object is ##dW=\vec F \cdot d\vec s##. The total work is the sum of all such terms. To find the sign of the work, you need to figure out when a dot product is positive and when it is negative. Any ideas?
 
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Thanks Kuruman,

So if the vectors of F & Δs are opposite (180) then the result is negative when using ΔW = |F| |Δs| cos 180.

I think i was getting confused between the direction of motion vs the change in position vector.

Thank you for your help!
 
More generally, ΔW = |F| |Δs| cosθ is negative when 90o < θ ≤ 180o.
Ben Harris said:
I think i was getting confused between the direction of motion vs the change in position vector.
Are the two different? The direction of motion is the same as the direction of the velocity vector. The velocity vector is defined as ##\vec v = \frac{d \vec s}{dt}## therefore ##\vec v## and ##d\vec s## point in the same direction.
 
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Ohhhhhh Yes!, that is correct... it will be negative!

Thank you for making me work for the answer :)
 
If the work done by gravity is negative the velocity of the meteor should decrease, shouldn't it?
 
nasu said:
If the work done by gravity is negative the velocity of the meteor should decrease, shouldn't it?
Yes.
 
kuruman said:
Yes.
The question was for the OP. :)
 
nasu said:
The question was for the OP. :)
Sorry, I was in autopilot. :sorry:
 

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