A SATELLITE ORBIT
You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth’s surface. (a) What speed, period, and radial acceleration will it have? (b) How much work must be done to the satellite to put it in orbit? (c) How much additional work would have to be done to make the satellite escape the earth? The earth’s radius and mass are given in Example 13.5 (Section 13.3).
IDENTIFY and SET UP:
The satellite is in a circular orbit, so we can use the equations derived in this section. In part (a), we first find the radius r of the satellite’s orbit from its altitude. We then calculate the speed v and period T from Eqs. (13.10) and (13.12) and the acceleration from a_{\mathrm{rad}}=v^{2} / r.In parts (b) and (c), the work required is the difference between the initial and final mechanical energies, which for a circular orbit is given by Eq. (13.13).
v=\sqrt{\frac{G m_{\mathrm{E}}}{r}} (13.10)
T=\frac{2 \pi r}{{ }v}=2 \pi r \sqrt{\frac{r}{G m_{\mathrm{E}}}}=\frac{2 \pi r^{3 / 2}}{\sqrt{G m_{\mathrm{E}}}} (13.12)
\begin{aligned}E &=K+U=\frac{1}{2} m v^{2}+\left(-\frac{G m_{\mathrm{E}} m}{r}\right) \\&=\frac{1}{2} m\left(\frac{G m_{\mathrm{E}}}{r}\right)-\frac{G m_{\mathrm{E}} m}{r} \\&=-\frac{G m_{\mathrm{E}} m}{2 r} \quad(\text { circular orbit })\end{aligned} (13.13)
EXECUTE:
(a) The radius of the satellite’s orbit is r = 6370 km + 300 km = 6670 km = 6.67 × 10^6 m. From Eq. (13.10), the orbital speed is
\begin{aligned}v &=\sqrt{\frac{G m_{\mathrm{E}}}{r}}=\sqrt{\frac{\left(6.67 \times 10^{-11}\mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(5.97 \times 10^{24} \mathrm{~kg}\right)}{6.67 \times 10^{6} \mathrm{~m}}} \\&=7730 \mathrm{~m} / \mathrm{s}\end{aligned}We find the orbital period from Eq. (13.12):
T=\frac{2 \pi r}{v}=\frac{2 \pi\left(6.67 \times 10^{6} \mathrm{~m}\right)}{7730 \mathrm{~m} / \mathrm{s}}=5420 \mathrm{~s}=90.4 \mathrm{~min}Finally, the radial acceleration is
a_{\mathrm{rad}}=\frac{v^{2}}{r}=\frac{(7730 \mathrm{~m} / \mathrm{s})^{2}}{6.67 \times 10^{6} \mathrm{~m}}=8.96 \mathrm{~m} / \mathrm{s}^{2}This is the value of g at a height of 300 km above the earth’s surface; it is about 10% less than the value of g at the surface.
(b) The work required is the difference between E_2, the total mechanical energy when the satellite is in orbit, and E_1, the total mechanical energy when the satellite was at rest on the launch pad. From Eq. (13.13), the energy in orbit is
\begin{aligned}E_{2} &=-\frac{G m_{\mathrm{E}} m}{2 r} \\&=-\frac{\left(6.67 \times 10^{-11}\mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(5.97 \times 10^{24}\mathrm{~kg}\right)(1000 \mathrm{~kg})}{2\left(6.67 \times 10^{6} \mathrm{~m}\right)} \\&=-2.98 \times10^{10} \mathrm{~J}\end{aligned}The satellite’s kinetic energy is zero on the launch pad (r = R_E), so
\begin{aligned}E_{1} &=K_{1}+U_{1}=0+\left(-\frac{G m_{\mathrm{E}} m}{R_{\mathrm{E}}}\right) \\&=-\frac{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(5.97 \times 10^{24} \mathrm{~kg}\right)(1000 \mathrm{~kg})}{6.37 \times 10^{6}\mathrm{~m}} \\&=-6.25 \times 10^{10} \mathrm{~J}\end{aligned}Hence the work required is
\begin{aligned}W_{\text {required }} &=E_{2}-E_{1}=\left(-2.98 \times 10^{10} \mathrm{~J}\right)-\left(-6.25 \times 10^{10} \mathrm{~J}\right) \\&=3.27 \times 10^{10} \mathrm{~J}\end{aligned}(c) We saw in part (b) of Example 13.5 that the minimum total mechanical energy for a satellite to escape to infinity is zero. Here, the total mechanical energy in the circular orbit is E_2 = -2.98 × 10^{10} J; to increase this to zero, an amount of work equal to 2.98 × 10^{10} J would have to be done on the satellite, presumably by rocket engines attached to it.
EVALUATE: In part (b) we ignored the satellite’s initial kinetic energy (while it was still on the launch pad) due to the rotation of the earth. How much difference does this make? (See Example 13.5 for useful data.)