Question 7.12: Goal Use gravitational potential energy to calculate the wor...
Goal Use gravitational potential energy to calculate the work done by gravity on a falling object.
Problem An asteroid with mass m = 1.00 × 10⁹ kg comes from deep space, effectively from infinity, and falls toward Earth. (a) Find the change in potential energy when it reaches a point 4.00 × 10⁸ m from Earth ( just beyond the Moon), assuming it falls from rest at infinity. In addition, find the work done by the force of gravity. (b) Calculate the speed of the asteroid at that point. (c) How much work would have to be done on the asteroid by some other agent so the asteroid would be traveling at only half the speed found in (b) at the same point?
Strategy Part (a) requires simple substitution into the definition of gravitational potential energy. To find the work done by the force of gravity, recall that the work done on an object by a conservative force is just the negative of the change in potential energy. Part (b) can be solved with conservation of energy, and part (c) is an application of the work–energy theorem.
(a) Find the change in potential energy and the work done by the force of gravity
Apply Equation 7.21:
PE = – G \frac{M_Em}{r} [7.21]
ΔPE = PE_f – PE_i = – \frac{G M_Em}{r_f} – (- \frac{G M_Em}{r_i})
= GM_Em( – \frac{1}{r_f} + \frac{1}{r_i})
Substitute known quantities. The asteroid’s initial position is effectively infinity, so 1/r_i is zero.
ΔPE = (6.67 × 10^{-11} kg^{-1}m^3/s^2)(5.98 × 10^{24} kg)
× (1.00 × 10^9 kg)( – \frac{1}{4.00 × 10^8 m} + 0)
ΔPE = – 9.97 × 10^{14} J
Compute the work done by the force of gravity:
W_{grav} = – ΔPE = 9.97 × 10^{14} J
(b) Find the speed of the asteroid when it reaches r_f = 4.00 × 10^8 m
Use conservation of energy:
ΔKE + ΔPE = 0
(\frac{1}{2}mv^2 – 0) – 9.97 × 10^{14} J = 0
v = 1.41 × 10^3 m/s
(c) Find the work needed to reduce the speed to 7.05 × 10² m/s (half the value just found) at this point.
Apply the work–energy theorem:
W = ΔKE + ΔPE
The change in potential energy remains the same as in part (a), but substitute only half the speed in the kinetic-energy term:
W = (\frac{1}{2}mv^2 – 0) – 9.97 × 10^{14} J
W = \frac{1}{2}(1.00 × 10^9 kg)(7.05 × 10^2 m/s)^2 – 9.97 × 10^{14} J
= – 7.48 × 10^{14} J
Remark The amount of work calculated in part (c) is negative because an external agent must exert a force against the direction of motion of the asteroid. It would take a thruster with a megawatt of output about 24 years to slow down the asteroid to half its original speed. An asteroid endangering Earth need not be slowed that much: A small change in its speed, if applied early enough, will cause it to miss Earth. Timeliness of the applied thrust, however, is important. By the time you can look over your shoulder and see the Earth, it’s already far too late, despite how these scenarios play out in Hollywood. Last minute rescues won’t work!