How does Equation 13.15 for the gravitational potential energy relate to our previous use of $U_{\text{G}}=mgy$ on a flat earth? FIGURE 13.14 shows an object of mass m located at height y above the surface of the earth. The object’s distance from the earth’s center is $r=R_{\text{e}}+y$ and its gravitational potential energy is

where, in the last step, we factored $R_{\text{e}}$ out of the denominator.

FIGURE 13.14 Gravity on a flat earth.

Suppose the object is very close to the earth’s surface $(y\ll R_{\text{e}})$. In that case, the ratio $y\text{/}{R}_{\text{e}}\ll 1$. There is an approximation you will learn about in calculus, called the binomial approximation, that says

As an illustration, you can easily use your calculator to find that $1\text{/}1.01=0.9901$, to four significant figures. But suppose you wrote $1.01=1+0.01$. You could then use the binomial approximation to calculate

You can see that the approximate answer is off by only 0.01%.

If we call $y\text{/}{R}_{\text{e}}=x$ in Equation 13.16 and use the binomial approximation, with $n=-1$, we find

Now the first term is just the gravitational potential energy $U_0$ when the object is at ground level $(y=0)$. In the second term, you can recognize $G{M}_{\text{e}}\text{/}{R}_{\text{e}}{}^2=g_{\text{earth}}$ from the definition of g in Equation 13.7. Thus we can write Equation 13.18 as

Although we chose $U_{\text{G}}$ to be zero when $r=\infty$, we are always free to change our minds. If we change the zero point of potential energy to be $U_0=0$ at the surface, which is the choice we made in Chapter 10, then Equation 13.19 becomes

We can sleep easier knowing that Equation 13.15 for the gravitational potential energy is consistent with our earlier “flat-earth” expression for the potential energy.

EXAMPLE 13.3 The speed of a satellite

A less-than-successful inventor wants to launch small satellites into orbit by launching them straight up from the surface of the earth at very high speed.

1. With what speed should he launch the satellite if it is to have a speed of $500\text{m}\text{/}\text{s}$ at a height of 400 km? Ignore air resistance.

2. By what percentage would your answer be in error if you used a flat-earth approximation?

MODEL Mechanical energy is conserved if we ignore drag.

VISUALIZE FIGURE 13.15 shows a pictorial representation.

FIGURE 13.15 Pictorial representation of a satellite launched straight up.

FIGURE 13.15 Full Alternative Text

SOLVE

1. Although the height is exaggerated in the figure, $400\text{km}=400\text{,}000\text{m}$ is high enough that we cannot ignore the earth’s spherical shape. The energy conservation equation $K_2+U_2=K_1+U_1$ is

   $$\frac{1}{2}m{v}_2{}^2-\frac{G{M}_{\text{e}}m}{R_{\text{e}}+y_2}=\frac{1}{2}m{v}_1{}^2-\frac{G{M}_{\text{e}}m}{R_{\text{e}}+y_1}$$

   where we’ve written the distance between the satellite and the earth’s center as $r=R_{\text{e}}+y$. The initial height is $y_1=0$. Notice that the satellite mass m cancels and is not needed. Solving for the launch speed, we have

   $$v_1=\sqrt{v_2{}^2+2G{M}_{\text{e}}\left(\frac{1}{R_{\text{e}}}-\frac{1}{R_{\text{e}}+y_2}\right)}=2770\text{ m}\text{/}\text{s}$$

   This is about 6000 mph, much less than the escape speed.

2. The calculation is the same in the flat-earth approximation except that we use $U_{\text{G}}=mgy$. Thus

   $$\begin{array}{l}\frac{1}{2}m{v}_2{}^2+mg{y}_2=\frac{1}{2}m{v}_1{}^2+mg{y}_1\\ v_1=\sqrt{v_2{}^2+2g{y}_2}=2840\text{ m}\text{/}\text{s}\end{array}$$

   The flat-earth value of $2840\text{m}\text{/}\text{s}$ is $70\text{m}\text{/}\text{s}$ too big. The error, as a percentage of the correct $2770\text{m}\text{/}\text{s}$, is

   $$\text{error}={\displaystyle \frac{70}{2770}}\times 100=2.5\%$$

REVIEW The true speed is less than the flat-earth approximation because the force of gravity decreases with height. Launching a rocket against a decreasing force takes less effort than it would with the flat-earth force of mg at all heights.