Work done against gravity with reducing gravity......

AI Thread Summary
To calculate the energy required to move a 100kg mass from the surface of Mars to a height of 8 x 10^6 m with non-linear gravity, one must consider the changing gravitational force as the object rises. The gravitational potential energy (GPE) equation typically assumes a constant gravitational pull, but for varying gravity, integration of the gravitational force law is necessary. The net force acting on the body is derived from Newton's law of gravitation, leading to an integral that represents the work done against gravity. The discussion emphasizes the importance of using the provided graph of gravitational force to accurately determine energy changes over the specified height. Understanding these concepts is crucial for solving the problem accurately.
SiriusFire
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Homework Statement
Calculate the energy required to move a 100kg mass from the surface of Mars to a height of 8 x10^6 m .
Relevant Equations
E = M x G x H
To do this I'm given a graph showing gravity reducing non-linearly from 3.7 N/kg at the surface to 0.7 N/kg at 8 x 10^6m. I believe that the equation for a fixed gravitaional pull is E = M x G x H - however with changing gravity as the object rises - how do you calculate this?
 
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SiriusFire said:
Homework Statement:: Calculate the energy required to move a 100kg mass from the surface of Mars to a height of 8 x10^6 m .
Relevant Equations:: E = M x G x H

To do this I'm given a graph showing gravity reducing non-linearly from 3.7 N/kg at the surface to 0.7 N/kg at 8 x 10^6m. I believe that the equation for a fixed gravitaional pull is E = M x G x H - however with changing gravity as the object rises - how do you calculate this?

Do you know the equation for GPE (gravitational potential energy) in the field of a spherical mass?

Hint: you can derive it from the gravitational force law.
 
Doesn't the GPE equation assume constant gravitational pull?
 
SiriusFire said:
Doesn't the GPE equation assume constant gravitational pull?
No. Do you know the gravitational force equation?

These days all you have to do is type "gravitational potential energy" into a search engine.
 
I assume that the Gravitational Force Equation isn't Pe = M g Δh then because when I type "gravitational potential energy" into a search engine that's what appears! And that always assumes constant g.
 
SiriusFire said:
Doesn't the GPE equation assume constant gravitational pull?
There is a simplified GPE equation which assumes that as an approximation, but it is not the (Newtonian) GPE eqution.
How does potential vary with distance from a spherically symmetric body for a force following an inverse square law, whether it be gravitational or electrostatic?
If you do not know, try integrating the force.
 
SiriusFire said:
I assume that the Gravitational Force Equation isn't Pe = M g Δh then because when I type "gravitational potential energy" into a search engine that's what appears! And that always assumes constant g.

Try this:

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html
 
OK Thanks - I'm struggling but I'll try from there!
 
SiriusFire said:
OK Thanks - I'm struggling but I'll try from there!
In case you don't know, Newton's law of gravitation says that we have a mutual force of attraction between two point masses, with the magnitude of the force given by:
$$F = \frac{Gm_1m_2}{r^2}$$
Where ##r## is the distance between the masses,and ##G## is the universal gravitational constant.
 
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... which you can integrate wrt r to find the potential.
 
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The work done on a body by the net force ##\vec F_{net}## acting on it, or the change in its kinetic energy, is given by:$$\int_\vec{s_1}^\vec{s_2}\vec F_{net}\cdot d\vec s$$
Where ##\vec s## is the displacement vector.
In your problem, the net force on the body is composed of only the gravitational pull the body feels because of the planet, and the displacement is straight upwards.
PeroK has given you the expression of the force. The integral simplifies to:
$$\int_{R_\text{mars}}^h-\frac{Gm_1m_2}{r^2}dr$$
This geometrically represents the area under the curve of the force against height in the graph you have from ##r=## the radius of mars, to ##r=h##.

It is unlikely that the problem wants you to find an approximation since the height is too big + he gave you a graph of the force, but here's how you can do it:
$$m_1a=\frac{Gm_1m_\text{mars}}{R_{mars}^2}\Leftrightarrow a=\frac{Gm_\text{mars}}{R_{mars}^2}$$
For heights close to the surface of mars, you can assume that the acceleration due the gravity of the planet is ##g_\text{mars}=a##, so the force your body feels is ##mg_\text{mars}##. Hence ##\Delta K=m_1g_\text{mars}h##.
 
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