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Center of earth Work problem (some calculus)

  • Thread starter Sammbuch
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Homework Statement



The force of attraction on an object below earth's surface is directly proportional to it's distance from earth's center.Find the work done in moving a weight of ##w## lb located ##a## miles below earth's surface up to the surface itself. Assume earth's radius is a constant r miles.

Homework Equations


##F = k*x##

The Attempt at a Solution




I would like someone to check this over. The work I have done so far (no pun intended). We are given that the force in magnitude is proportional to the distance, or ##F = k*x## where ##x## is the distance from the center of earth to the object. Now the question is a little confusing, how to find k. I think we can assume they mean at the surface of earth it weighs ##w ## lbs, therefore ##w = k* r## and ##k = w/r##, where ##r## is the radius of earth. We want the work for ##a## miles below the surface of earth.

We have the integral:
##Work=\int\limits_{r-a}^{r}\frac{w}{r}xdx= \frac{w(2ar-a^2)}{2r}##

This does not agree with book answer
##\frac{w}{2}(2ar-a^2)##

What did I do wrong?

Also as a check if this makes sense, I found the total work to move a mass of weight w lbs from the center of the earth to the surface of the earth is :
##\int\limits_{0}^{r}\frac{w}{r}xdx= \frac {1}{2}wr##

This implies if you drilled a tunnel through the surface of the earth all the way to the opposite surface and dropped an object of weight w lbs, what would be the work done on the mass? ##2*1/2wr= wr##, where r is the radius of earth.
 
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Answers and Replies

  • #2
TSny
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Your work looks correct to me. Note that the book's answer does not have the correct dimensions for work.

For your tunnel example, does the work for the first half of the trip (from surface to center) have the same sign as the second half (from center to surface at other side)?
 
  • #3
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that makes sense, dimensional analysis gives me ##mi*lb## or ##ft*lb## by multiplying by 5280.

As for your second reply does negative work make sense, i guess it does if you account for direction and choose up as positive.
If you allow for negative work, then the work should come out to zero?

But maybe there is an absolute version of total work, like we have total distance versus displacement. Then it will be twice the amount of work.
 
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  • #4
TSny
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Work has the dimensions of force times distance.

If an object moves in the same direction as the direction of the force, then the work is positive. If it moves in a direction opposite of the direction of the force, then the work is negative.

I don't think there is any usefulness in defining an "absolute version" of work. Work is defined so that the net work done on a particle is equal to the change in kinetic energy of the particle. When you drop a particle into the tunnel, what is the overall change in kinetic energy from the initial drop point to the point at the surface on the other side of the earth? You want the total work to equal this change in kinetic energy energy.
 
  • #5
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ok so finding the net work using the gravity approach should come out to the same answer using the kinetic energy approach.
So you will have
##\int\limits_{-r}^{r}\frac{w}{r}xdx =0##
and the kinetic energy will be the same 0 speed at top and bottom ##\frac{1}{2}mv^2## gives us zero when initial velocity is zero.
i think kinetic energy always has to be positive.
 
  • #6
TSny
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Yes, kinetic energy is never negative. But, change in kinetic energy can be negative.
 
  • #7
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that makes sense
 
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  • #8
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It is interesting that you can find the total or net work done using the much easier kinetic energy approach.
Due to total energy being conserved, it's like a yo yo (harmonic oscillator). The mass starts at speed of zero, and accelerates as it goes down the tunnel because the earth is pulling on it, and reaches a maximum speed at the center of the earth. Then the speed slows down as it goes to the other opposite surface of the earth, until it reaches zero speed.
 
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  • #9
TSny
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Yes. You will learn that work and energy are powerful concepts that often provide an easier way to analyze a problem compared to using force and acceleration.
 

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