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

AI Thread Summary
The discussion revolves around determining whether the work done by gravity on an approaching meteor is positive or negative. The original poster calculated the work using the formula W(1->2) = integral (1->2) F(gravity) dot dr and concluded that it should be positive, as GMem/r2 is greater than GMem/r1. However, further clarification revealed that the sign of the work depends on the direction of the force and the change in position vector, leading to the conclusion that if the vectors are opposite, the work is negative. This negative work indicates that the meteor's velocity decreases as it approaches Earth. Ultimately, the consensus is that the work done by gravity on the meteor is negative, resulting in a decrease in its velocity.
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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