Set 7 Problem number 7

A tower begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 1000 kilometers further from the center.

How much force would an individual of mass 91 kg need to exert against gravity at the beginning and at the end of the climb, in order to climb this tower?
If the average force exerted was equal to the average of the initial and final force, how much energy would be required to climb the tower?
If the individual was reasonably well-conditioned and capable of an average sustained power output of 1.11 watt/kg for 8 hours per day, how many days would be required to climb the tower?

Solution

The force required at the surface of the Earth is force to rise at constant velocity against gravity at Earth surface:

F = 91 kg (9.8 m/s ^ 2) = 891.8 Newtons.

At altitude 1000 kilometers above the surface, the ratio of radii will be

radius ratio: (6400 + 1000) km / 6400 km,

so the gravitational field will have magnitude

grav field at altitude: g = (9.8 m/s ^ 2) / [(6400+ 1000)/6400] ^ 2

resulting in a force of

gravitational force (weight) at 1000 km altitude = 91 kg (9.8 m/s ^ 2) / [(6400+ 1000)/6400] ^ 2 = 667.05 Newtons.

The average of the two forces is

average of forces = 779.425 Newtons.

[Note that since force is not a linear function of distance this average of the two forces will only be an approximation, and perhaps not a very good one, to the average force].

Exerted over a distance of 1000 km = 1000000 meters, this force would require energy equal to the work done

energy required = ( 779.425 Newtons)( 1000000 meters) = 7.79425E+08 Joules.

An average power output of 1.11 Watts/kilogram implies that the individuals average power output is

average power output of individual = 101.01 Watts.

In 8 hours, the individual would therefore produce

energy in 8 hour working day = ( 101.01 Joules/second) (8 hours) (3600 seconds/hour) = 2909088 Joules.

To produce the required energy would therefore require

time required = 7.79425E+08 Joules/( 2909088 Joules/day) = 267.9 days.

Generalized Solution

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 (R / r1) ^ 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 (R / r1) ^ 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 (R / r1) ^ 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 = R and r = 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

approx ave force = (F (r(n) ) + F (r(n+1) ) / 2

= (m g * (R / r(n))^2 + m g * ( R / r(n+1) )^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. That is, the gravitational force is conservative.

The work done to 'lift' the object is therefore equal to the potential energy increase of the planet-object system.

Explanation in terms of Figure(s)

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 than 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 to 'raise' the object is in the direction of the corresponding displacement, the work required to move from distance r1 to r2 is the product of the distance (r2 - r1) traveled and by the average force Fave. This work is equal to the potential energy increase of the object.