Finding thing due to gravitation

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The discussion revolves around calculating the effects of air drag on Earth's orbital radius, specifically focusing on changes in orbital velocity, kinetic energy, potential energy, and work done by air resistance. Participants clarify that while air drag is negligible for Earth, it can affect satellites, leading to a decrease in orbital radius. The conversation includes deriving formulas for changes in velocity and energy, emphasizing the importance of understanding central forces and angular momentum conservation. There is also a debate on the relevance of potential energy equations in this context, with some participants seeking clarity on concepts like kinetic and potential energy. Ultimately, the group arrives at a solution for calculating the changes in energy due to the hypothetical scenario presented.
Raghav Gupta
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Homework Statement


Assume that the orbital radius of the Earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
Then what is change in orbital velocity, change in kinetic energy, change in potential energy and work done by air resistance?

Homework Equations


F = Gm1m2/r2
Kinetic energy = 1/2 mv2
Potential energy = mgh

The Attempt at a Solution


Work done by air resistance is Gm1m2Δr/r2 ?
 
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Hello Raghav,

No A, B, C, D this time ? Where does this exercise come from ?
Can you help me imagine the situation ? This about an Earth moving around the sun in air instead of in a vacuum ?

My impression (more like a proven hypothesis) was always that central forces conserve angular momentum and I don't see anything non-central in this.

Are you assuming circular orbits ? There is no law the orbit should be circular, as far as I know.

(In fact Earth orbit excentricity is some 3 million km, much more than Earth radius !)
 
Raghav Gupta said:

Homework Statement


Assume that the orbital radius of the Earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
Then what is change in orbital velocity, change in kinetic energy, change in potential energy and work done by air resistance?

Homework Equations


F = Gm1m2/r2
Kinetic energy = 1/2 mv2
Potential energy = mgh
Potential energy of what? Is this equation relevant for the problem?

Raghav Gupta said:

The Attempt at a Solution


Work done by air resistance is Gm1m2Δr/r2 ?
How do you get the work of a force? Is Gm1m2/r2 the air resistance?
 
BvU said:
My impression (more like a proven hypothesis) was always that central forces conserve angular momentum and I don't see anything non-central in this.

Well, not "air" resistance, but the space is not empty, there are molecules and dust particles which interact with the Earth and exert some force of resistance, reducing its speed. But such forces are opposite to the velocity. The velocity is not parallel to the radius of the orbit. The resistive force has got torque and changes the angular momentum.
 
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Well this was a matrix match type question
And I don't know how to make a matrix with latex.
The second column was
## \frac{-GmMΔR}{R^2} ##

## \frac{-GmMΔR}{2R^2} ##

## \frac{GmMΔR}{2R^2} ##

## \frac{ΔR}{2} \sqrt{\frac{GM}{R^3}} ##

This exercise was of a practice, not directly from an exam.
Yeah, there is no air in space and orbit is approximately elliptical
ehild said:
Potential energy of what? Is this equation relevant for the problem?

How do you get the work of a force? Is Gm1m2/r2 the air resistance?
Well I was looking potential energy in Wikipedia and they say it is because of some field.
They have written for capacitor ½ CV2 is potential energy. Why it is called potential energy! Because it is of electric field?
Whenever someone calls kinetic energy ,the first thing comes to my mind is ½mv2.
But I know it is energy due to motion.
I have not applied it in other scenarios.

The work expression is wrong for air resistance, I know. I was not able to imagine it properly.
Work done = Force * displacement cosθ
 
I think one can use dimensional analysis to match options only helping for orbital velocity.
But what is the correct logical way?
 
Raghav Gupta said:
Well I was looking potential energy in Wikipedia and they say it is because of some field.
They have written for capacitor ½ CV2 is potential energy. Why it is called potential energy! Because it is of electric field?
Whenever someone calls kinetic energy ,the first thing comes to my mind is ½mv2.
But I know it is energy due to motion.

The work expression is wrong for air resistance, I know. I was not able to imagine it properly.
Work done = Force * displacement cosθ

Is it real that you were not taught about kinetic and potential energy in the school?
About conservative forces? Energy of a planet orbiting about the Sun? Kepler's Laws?
Raghav Gupta said:
I think one can use dimensional analysis to match options only helping for orbital velocity.
But what is the correct logical way?
You can assume circular orbit and assume that the one with smaller radius is also a stable orbit. Knowing the radius, you get the speed with comparing the centripetal force to the force of gravity.
The potential energy around a central mass M is -GmM/R.
Knowing potential energy and kinetic energy on both orbits, you can find the change of the total energy, due to the "air resistance".
 
I was taught about it in school.
But I have not applied it in space, means I have applied it for objects on ground level.
What is the difference when we go in space about these energies?
Don't know energy of a planet orbiting sun.
Is it Gm1m2/R where R is distance and m1 and m2 masses of sun and earth?
Kepler laws I know.

So Mv2/R = GMm/R2 M is mass of Earth and m of sun.
 
Raghav Gupta said:
Assume that the orbital radius of the Earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
Presumably this is meant to be the orbital radius of an Earth-orbiting satellite, yes?
 
  • #10
gneill said:
Presumably this is meant to be the orbital radius of an Earth-orbiting satellite, yes?
Yes I think so. As Earth is so big and air drag on it does not make sense.
 
  • #11
So orbital velocity = ## \sqrt{\frac{GM}{R}} ##
How to calculate Δv ?
Should I differentiate with respect to M or R?
But mass of Earth and radius of Earth are constant.
If we take R as satellite orbital radius.
Then ## \frac{dv}{dR} = \frac{-1}{2R} \sqrt{\frac{GM}{R}} ##
so
## Δv = \frac{-ΔR}{2} \sqrt{\frac{GM}{R^3}} ##
But in options - sign is not there.
 
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  • #12
Okay, got it,
Here dr = R -ΔR - R = -ΔR.
So I will get correct option.
Now I can easily use K.E = 1/2mv2 and get things.
Also for potential energy = -GMm/R
Work energy theorem for air drag.
It would be a lot of latex to show but I have got it.
Thanks BvU, ehild and gneill.
I'm now on a long break.
 
  • #13
You deserve it !
 
  • #14
I wanted to try to fix up the title "Finding thing due to gravitation", but I'm still not sure what to change it to...
 
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