Problem: A person of mass 52 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 climbs to a position 900 kilometers further from the center. How much force does the individual exert against gravity at the beginning and at the end of the climb? If the average force exerted was equal to the average of the initial and final force, how much energy would be required? At an average power output of .6500 watt/kg for 8 hours per day, how many days would be required?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: The force required at the surface of the Earth is 52 kg (9.8 m/s ^ 2) = 509.6 Newtons.
At altitude 900 kilometers above the surface, the ratio of radii will be (6400 + 900) km / 6400 km, so the gravitational field will have magnitude (9.8 m/s ^ 2) / [(6400+ 900)/6400] ^ 2 resulting in a force of 52 kg (9.8 m/s ^ 2) / [(6400+ 900)/6400] ^ 2 = 391.6 Newtons.
The average of these forces is 450.6 Newtons. Exerted over a distance of 900 km = 9.000 * 10 ^ 5 meters, this force would require ( 450.6 Newtons)( 9.000 * 10 ^ 5 meters) = 4.055 * 10 ^ 7 Joules of energy.
An average power output of .6500 Watts/kilogram implies that the individuals average power output is 33.8 Watts.
In an 8 hour day, the individual would therefore produce ( 33.8 Joules/second) (8 hours) (3600 seconds/hour) = 9.734 * 10 ^ 5 Joules (rate is therefore 9.734 * 10 ^ 5 Joules per day).
To produce the required energy would therefore require 4.055 * 10 ^ 7 Joules/( 9.734 * 10 ^ 5 Joules/day) = 41.65 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.
.
.
.
.
.
.
.
.
.
.
Figure Description: The figure below shows a poorly constructed tower rising from Earth (the black dot at the center), a blue sphere concentric with Earth and having radius r1, and a similar green sphere of radius r2. The forces F1 and F2 required for mass m to climb the tower at constant speed at distances r1 and r2, respectively, from the center of the Earth are depicted by red vectors, and the approximate average force Fave is also depicted by a red vector.
The gravitational effect of Earth is spread over a smaller area on the smaller blue sphere then on the larger green sphere, resulting in a larger gravitational field at the blue sphere. This results in a greater gravitational acceleration toward the Earth when the object is at distance r1 than at r2, and hence a greater force necessary to overcome this acceleration.
The force does not fall off linearly, but as an inverse square. The average force is therefore not equal to the average of the two forces F1 and F2, but if the forces do not differ by too much we get a reasonable approximation to the average force.
Since the force required is in the direction of motion, the work required to move from distance r1 to r2 is the distance (r2 - r1) traveled multiplied by the average force, and is equal to the potential energy increase of the object.
