13.3 Gravitational potential energy and total energy  (Page 4/6)

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As noted earlier, we see that $U\to 0\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}r\to \infty$ . If the total energy is zero, then as m reaches a value of r that approaches infinity, U becomes zero and so must the kinetic energy. Hence, m comes to rest infinitely far away from M . It has “just escaped” M . If the total energy is positive, then kinetic energy remains at $r=\infty$ and certainly m does not return. When the total energy is zero or greater, then we say that m is not gravitationally bound to M .

On the other hand, if the total energy is negative, then the kinetic energy must reach zero at some finite value of r , where U is negative and equal to the total energy. The object can never exceed this finite distance from M , since to do so would require the kinetic energy to become negative, which is not possible. We say m is gravitationally bound    to M .

We have simplified this discussion by assuming that the object was headed directly away from the planet. What is remarkable is that the result applies for any velocity. Energy is a scalar quantity and hence [link] is a scalar equation—the direction of the velocity plays no role in conservation of energy. It is possible to have a gravitationally bound system where the masses do not “fall together,” but maintain an orbital motion about each other.

We have one important final observation. Earlier we stated that if the total energy is zero or greater, the object escapes. Strictly speaking, [link] and [link] apply for point objects. They apply to finite-sized, spherically symmetric objects as well, provided that the value for r in [link] is always greater than the sum of the radii of the two objects. If r becomes less than this sum, then the objects collide. (Even for greater values of r , but near the sum of the radii, gravitational tidal forces could create significant effects if both objects are planet sized. We examine tidal effects in Tidal Forces .) Neither positive nor negative total energy precludes finite-sized masses from colliding. For real objects, direction is important.

How far can an object escape?

Let’s consider the preceding example again, where we calculated the escape speed from Earth and the Sun, starting from Earth’s orbit. We noted that Earth already has an orbital speed of 30 km/s. As we see in the next section, that is the tangential speed needed to stay in circular orbit. If an object had this speed at the distance of Earth’s orbit, but was headed directly away from the Sun, how far would it travel before coming to rest? Ignore the gravitational effects of any other bodies.

Strategy

The object has initial kinetic and potential energies that we can calculate. When its speed reaches zero, it is at its maximum distance from the Sun. We use [link] , conservation of energy, to find the distance at which kinetic energy is zero.

Solution

The initial position of the object is Earth’s radius of orbit and the intial speed is given as 30 km/s. The final velocity is zero, so we can solve for the distance at that point from the conservation of energy equation. Using ${R}_{\text{ES}}=1.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}$ and ${M}_{\text{Sun}}=1.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}$ , we have

$\begin{array}{c}\frac{1}{2}m{v}_{1}^{2}-\frac{GMm}{{r}_{1}}=\frac{1}{2}m{v}_{2}^{2}-\frac{GMm}{{r}_{2}}\hfill \\ \\ \\ \\ \phantom{\rule{1em}{0ex}}\frac{1}{2}\overline{)m}\left(3.0×{10}^{3}\text{m/s}{\right)}^{2}-\frac{\left(6.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m/kg}}^{2}\right)\left(1.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\overline{)m}}{1.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}}\hfill \\ \\ \phantom{\rule{4em}{0ex}}=\frac{1}{2}\overline{)m}{0}^{2}-\frac{\left(6.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m/kg}}^{2}\right)\left(1.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\overline{)m}}{{r}_{2}}\hfill \end{array}$

where the mass m cancels. Solving for ${r}_{2}$ we get ${r}_{2}=3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}$ . Note that this is twice the initial distance from the Sun and takes us past Mars’s orbit, but not quite to the asteroid belt.

Significance

The object in this case reached a distance exactly twice the initial orbital distance. We will see the reason for this in the next section when we calculate the speed for circular orbits.

Check Your Understanding Assume you are in a spacecraft in orbit about the Sun at Earth’s orbit, but far away from Earth (so that it can be ignored). How could you redirect your tangential velocity to the radial direction such that you could then pass by Mars’s orbit? What would be required to change just the direction of the velocity?

You change the direction of your velocity with a force that is perpendicular to the velocity at all points. In effect, you must constantly adjust the thrusters, creating a centripetal force until your momentum changes from tangential to radial. A simple momentum vector diagram shows that the net change in momentum is $\sqrt{2}$ times the magnitude of momentum itself. This turns out to be a very inefficient way to reach Mars. We discuss the most efficient way in Kepler’s Laws of Planetary Motion .

Summary

• The acceleration due to gravity changes as we move away from Earth, and the expression for gravitational potential energy must reflect this change.
• The total energy of a system is the sum of kinetic and gravitational potential energy, and this total energy is conserved in orbital motion.
• Objects must have a minimum velocity, the escape velocity, to leave a planet and not return.
• Objects with total energy less than zero are bound; those with zero or greater are unbounded.

Conceptual questions

It was stated that a satellite with negative total energy is in a bound orbit, whereas one with zero or positive total energy is in an unbounded orbit. Why is this true? What choice for gravitational potential energy was made such that this is true?

It was shown that the energy required to lift a satellite into a low Earth orbit (the change in potential energy) is only a small fraction of the kinetic energy needed to keep it in orbit. Is this true for larger orbits? Is there a trend to the ratio of kinetic energy to change in potential energy as the size of the orbit increases?

As we move to larger orbits, the change in potential energy increases, whereas the orbital velocity decreases. Hence, the ratio is highest near Earth’s surface (technically infinite if we orbit at Earth’s surface with no elevation change), moving to zero as we reach infinitely far away.

Problems

Find the escape speed of a projectile from the surface of Mars.

5000 m/s

Find the escape speed of a projectile from the surface of Jupiter.

What is the escape speed of a satellite located at the Moon’s orbit about Earth? Assume the Moon is not nearby.

1440 m/s

(a) Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-to-center distance of 15.0 cm. (b) Assuming that they are both initially at rest relative to each other in deep space, use conservation of energy to find how fast will they be traveling upon impact. Each sphere has a radius of 5.10 cm.

An average-sized asteroid located $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{km}$ from Earth with mass $2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}\phantom{\rule{0.2em}{0ex}}\text{kg}$ is detected headed directly toward Earth with speed of 2.0 km/s. What will its speed be just before it hits our atmosphere? (You may ignore the size of the asteroid.)

11 km/s

(a) What will be the kinetic energy of the asteroid in the previous problem just before it hits Earth? b) Compare this energy to the output of the largest fission bomb, 2100 TJ. What impact would this have on Earth?

(a) What is the change in energy of a 1000-kg payload taken from rest at the surface of Earth and placed at rest on the surface of the Moon? (b) What would be the answer if the payload were taken from the Moon’s surface to Earth? Is this a reasonable calculation of the energy needed to move a payload back and forth?

a. $5.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$ ; b. $-5.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$ ; No. It assumes the kinetic energy is recoverable. This would not even be reasonable if we had an elevator between Earth and the Moon.

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Kidus   By Lakeima Roberts       By Jordon Humphreys