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## Q. 7.11

A NEAR-EARTH ASTEROID

GOAL Use gravitational potential energy to calculate the work done by gravity on a falling object.

PROBLEM An asteroid with mass $m=1.00 \times 10^9 \mathrm{~kg}$ comes from deep space, effectively from infinity, and falls toward Earth. (a) Find the change in potential energy when it reaches a point $4.00 \times 10^8 \mathrm{~m}$ from the center of the Earth (just beyond the orbital radius of the Moon). In addition, find the work done by the force of gravity. (b) Calculate the asteroid’s speed at that point, assuming it was initially at rest when it was arbitrarily far away. (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.

## Verified Solution

(a) Find the change in potential energy and the work done by the force of gravity.

Apply Equation 7.22:

$PE = -G \frac{M_E m}{r}$    [7.22]

\begin{aligned}&\Delta P E=P E_f-P E_i=-\frac{G M_E m}{r_f}-\left(-\frac{G M_E m}{r_i}\right) \\&=G M_E m\left(-\frac{1}{r_f}+\frac{1}{r_i}\right)\end{aligned}

Substitute known quantities. The asteroid’s initial position is
effectively infinity, so $1 / r_i$ is zero:

\begin{aligned}\Delta P E=&\left(6.67 \times 10^{-11} \mathrm{~kg}^{-1} \mathrm{~m}^3 / \mathrm{s}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right) \\& \times\left(1.00 \times 10^9 \mathrm{~kg}\right)\left(-\frac{1}{4.00 \times 10^8 \mathrm{~m}}+0\right)\end{aligned}

$\Delta P E=-9.97 \times 10^{14} \mathrm{~J}$

Compute the work done by the force of gravity:

$W_{\text {grav }}=-\Delta P E=9.97 \times 10^{14} \mathrm{~J}$

(b) Find the speed of the asteroid when it reaches $r_f=4.00 \times 10^8 \mathrm{~m}$.

Use conservation of energy:

\begin{aligned}&\Delta K E+\Delta P E=0 \\&\left(\frac{1}{2} m v^2-0\right)-9.97 \times 10^{14} \mathrm{~J}=0 \\&v=1.41 \times 10^3 \mathrm{~m} / \mathrm{s}\end{aligned}

(c) Find the work needed to reduce the speed to $7.05 \times 10^2 \mathrm{~m} / \mathrm{s}$ (half the value just found) at this point.

Apply the work-energy theorem:

$W=\Delta K E+\Delta P E$

The change in potential energy remains the same as in part (a), but substitute only half the speed in the kinetic-energy term:

\begin{aligned}W &=\left(\frac{1}{2} m v^2-0\right)-9.97 \times 10^{14} \mathrm{~J} \\W &=\frac{1}{2}\left(1.00 \times 10^9 \mathrm{~kg}\right)\left(7.05 \times 10^2 \mathrm{~m} / \mathrm{s}\right)^2-9.97 \times 10^{14} \mathrm{~J} \\&=-7.48 \times 10^{14} \mathrm{~J}\end{aligned}

REMARKS 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 an astronaut on the asteroid can look over his 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!