Problem: A person of mass 72 kg begins climbing a very high tower. The tower begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 3300 kilometers further from the center. For each of the first three 1100 kilometer segments, determine the average of the initial and final forces for the segment. Give the total work required for each segment, based on the average of the initial and final forces for the segment. At an average power output of .9600 watt/kg for 8 hours per day, how many days would be required to make the 3300 kilometer climb?
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Solution: The force required at the surface of the Earth is 72 kg (9.8 m/s ^ 2) = 705.6 Newtons.
At altitude 3300 kilometers above the surface, the ratio of radii will be (6400 + 1100) km / 6400 km, so the gravitational field will have magnitude (9.8 m/s ^ 2) / [(6400+ 1100)/6400] ^ 2 resulting in a force of 72 kg (9.8 m/s ^ 2) / [(6400+ 1100)/6400] ^ 2 = 513.8 Newtons.
The average of these forces multiplied by the 1100 km = 1.100 * 10 ^ 6 meters yields work 6.706 * 10 ^ 7 Joules.
At altitude 2( 1100) km, gravity exerts a force of 72 kg (9.8 m/s ^ 2) / [(6400+2* 1100)/6400] ^ 2 = 284.5 Newtons. Exerted over a distance of 1100 km = 1.100 * 10 ^ 6 meters, the average of this force and the 513.8 Newtons required at the beginning of this segment would require 6.706 * 10 ^ 7 Joules of energy.
A similar calculation at altitude 3( 1100) kilometers yields a force of 123.8 Newtons, which will be averaged with 284.5 Newtons to obtain the approximate average force for this segment. Multiplying by the distance 1100 km covered on this segment we determined that the work required is 2.245 * 10 ^ 7 Joules.
The total work over all three segments is therefore 3.586 * 10 ^ 6 Joules. An average power output of .9600 Watts/kilogram applies that the individuals average power output is 69.12 Watts. In 8 hours, the individual would therefore produce ( 69.12 Joules/second) (8 hours) (3600 seconds/hour) = 1.990 * 10 ^ 6 Joules per day.
To produce the required energy would therefore require 3.586 * 10 ^ 6 Joules/( 1.990 * 10 ^ 6 Joules/day) = 1.802 days.
Generalized Response: If a planet of radius R has gravitational field g at its surface, then the field at distance r1 from its center is
field at distance r1 = g (r1 / R) ^ 2.
To move a mass m away from the planet, when that mass is at distance r1, will therefore require force
F ( r1 ) = m * gravitational field at distance r1 = m g (r1 / R) ^ 2.
Since the gravitational force is toward the center of the planet, the force required to move the mass will be directed away from the planet. The work done in moving the mass a short distance `dr directly away from the planet, moving the object from distance r1 to distance r1 + `dr, will therefore be approximately
work = force * distance = F ( r1 ) * `dr = m g (r1 / R) ^ 2 * `dr.
To the extent that `dr is small the actual force experienced will vary little from F ( r1 ) and this approximation will be good. If we divide the total distance from R to r1 into increments `dr which are small enough that the approximation over every increment is good, we can achieve a good approximation to the work done. (Calculus-based Physics students note that the limit of this process gives us the integral of F ( r1 ) dr, or m g / R^2 * r^2 `dr, between limits R and r1).
If we wish to approximate the work using a small number of increments we divide the distance into such increments. If increment number n goes from distance r(n) to distance r(n+1), then we have forces F( r(n) ) = m g ( r(n) / R ) ^ 2 and F( r(n+1) ) = m g ( r(n+1)^ / R) ^ 2 at the two extreme points of the increment, giving approximate average force
ave force = (F (r(n) ) + F (r(n+1) ) / 2 = (m g * (r(n) / R)^2 + m g * (r(n+1) / R)^2) / 2 = m g / R * (r(n)^2 + r(n+1)^2) / 2.
Adding up all such contributions gives an approximation to the total work. To the extent that the force does not vary too drastically over each increment, the approximation will be good.
The work done to increase the distance can of course be regained in the form of kinetic energy by allowing the object to fall back toward the planet. The work done is therefore equal to the potential energy increase of the planet-object system.
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Figure Description: The figure below shows a graph of the gravitational force vs. distance for a certain object, from a distance of 1 Earth radius to 4 Earth radii. The approximate average force between 1 and 2 Earth radii is indicated by a red vertical arrow. This average is just the average of the forces at 1 and 2 Earth radii, and as can be readily seen by visualizing the actual force curve, is significantly higher than the actual average force.
The distance from 1 to 2 Earth radii is also indicated by a red arrow.
The approximate work done in moving from 1 to 2 Earth radii is the product of the force (vertical red arrow) and the distance (horizontal red arrow), and is therefore identical to the area (height * width) of the red rectangle whose height and width match the arrows representing force and distance.
The work required to move from 2 to 3, and from 3 to 4 Earth radii can be similarly represented. The corresponding rectangles have been shown.
