A person of mass 66 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 3600 kilometers further from the center.
The first segment extends from the surface of Earth to altitude 1200 km--from distance 6400 km from Earth's center to (6400 + 1200) km.
The second segment extends from (6400 + 1200) km to (6400 + 2* 1200) km:
The third segment extends from (6400 + 2 * 1200) km to (6400 + 3 * 1200 km):
The approximate total work over all three segments is therefore the sum 1.289232E+09 Joules of the work estimates from the three segments. Note that this approxmation is smaller than the we would obtain from using just the forces at the surface and the 3 * 1200 km altitudes. Increasing numbers of segments would result in smaller approximations, which would approach a limit equal to the actual work required.
An average power output of 1.35 Watts/kilogram implies that the individual's average power output is 89.1 Watts.
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.
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.
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