9.8 Universal gravitation
Imagine that you launch a cannonball straight out of a cannon; it goes a bit, then lands. What happens if you shoot it with a bit more velocity? What if you shoot it really fast? You could shoot it with such a velocity that the ground is dropping out from under it at the same rate that it is falling—it would never land! The cannonball would now be in orbit, as shown below.
Isaac Newton [Public domain], via Wikimedia Commons
In the 17th century, Newton discovered that gravity is the force that keeps objects in orbit, and published it in Principia Mathematica. The implications of this were revolutionary—the same force that keeps things to the ground here on Earth governs celestial bodies! Gravity acts between any two bodies with mass; they are attracted towards each other. The magnitude of this force is given by:
\[
F^G = \frac{Gm_1m_2}{r^2} \tag{9.8}
\]
where \(m_1\) and \(m_2\) are the masses of the two objects, \(r\) is the distance between them, and \(G\) is the gravitational constant. Newton did not know what the value of \(G\) was; it was determined for the first time in the 1790s by Henry Cavendish. The value of “Big G” in SI units is \(G = 6.67 \times 10^{-11}\) N · m2 ⁄ kg2.
When you are near the surface of Earth, the distance from you to the center of Earth is the planet’s radius \(R_\oplus = 6.371 \times 10^6\ \textrm{m}\). The mass of Earth is \(M_\oplus = 5.972 \times 10^24\ \textrm{kg}\). If you have a mass \(m\), this means the force of gravity Earth exerts on you is
\[
\begin{align*}
F^G &= \frac{GmM_\oplus}{R_\oplus^2} \\
&= m\left(9.81\ \textrm{m/s}^2\right) \\
&= mg
\end{align*}
\]
This is how we know the value of \(g\), the strength of Earth’s gravitational field. In fact, you can find out the strength of Earth’s gravitational field even if you are not close to the surface: use \(R_\oplus + h\) for any height \(h\) above the surface.
Furthermore, you can find the strength of gravity on the surface of any planet (or moon, or asteroid large enough to be spherical, etc.), as long as you know that planet’s mass and radius:
\[
g_\textit{planet} = \frac{G M_\textit{planet}}{R_\textit{planet}^2}
\]
Example 9.12
Some believe that the positions of the planets on particular days can affect their lives; many people justify this with gravity. The Sun is the most massive object in the solar system. It has a mass of \(m_\odot = 1.98 \times 10^{30}\ \textrm{kg}\) and is a distance \(r_{S-E} = 1.49 \times 10^{11}\ \textrm{m}\) away. Jupiter, the most massive planet in the solar system, has a mass \(m_J = 1.89 \times 10^{27}\ \textrm{kg}\). It is 588 million kilometers from us \(\left(r_{J-E} = 5.88 \times 10^{11}\ \textrm{m}\right)\). Let’s say you have a mass \(m_h = 70\ \textrm{kg}\).
- What is the force of gravity of the Sun pulling on you?
- What is the force of gravity of Jupiter pulling on you?
- What is your weight (the force of gravity of Earth pulling on you)?
- How many times stronger is the Sun pulling on you than Jupiter pulls on you? (Find the ratio of the Sun’s pull on you to Jupiter’s)
- How many times stronger is the Earth pulling on you than the Sun pulls on you?
Calculate using Newton’s expression for general gravitation:
\[
\begin{align*}
F_{S-h} &= G\frac{m_Sm_h}{r_{S-E}^2} & F_{E-h} &= m_hg \\
&= 0.417\ \textrm{N} & &= 686\ \textrm{N} \\
F_{J-h} &= G\frac{m_Jm_h}{r_{J-E}^2} & \hookrightarrow F_{S-h} &= (1.63 \times 10^4)F_{J-h} \\
&= 2.55 \times 10^{-5}\ \textrm{N} & \hookrightarrow F_{E-h} &= (1.65 \times 10^3)F_{S-h}
\end{align*}
\]
So, Earth is pulling on us much more strongly than the Sun, and even more strongly than the most massive planet in the solar system. You may verify for yourself that the same holds for any other body in the solar system, such as the other planets or our moon.
flic.kr/p/5P143U
Example 9.13
A satellite goes around the earth in a circular orbit, 184 km above Earth’s surface. Earth’s mass is \(5.97 \times 10^{24}\ \textrm{kg}\) and its radius is \(6.37 \times 10^6\ \textrm{m}\). What is the speed of the satellite?
We are dealing with circular motion and the only force acting is gravity, which is pointing towards the center of the circle. I’ll call the radius of the earth \(R\), and the height of the satellite above the surface \(h\). Note that the radius of the satellite’s path (the distance from the satellite to the center of the earth) is \(r = R + h\). Apply Newton’s second law:
\[
\begin{align*}
F^G &= m_sa_c \\
G\frac{m_sM_e}{r^2} &= m_s\frac{v^2}{r} \\
G\frac{m_sM_e}{(R + h)^2} &= m_s\frac{v^2}{R + h} \\
v &= \sqrt{G\frac{M_e}{R + h}} \\
&= 7.80 \times 10^3\ \textrm{m/s} \\
&= 7.80\ \textrm{km/s}
\end{align*}
\]
Example 9.14
Jupiter has many moons, but four of them are easily visible with even a very inexpensive telescope. These moons are Io, Europa, Ganymede and Callisto. They are called the Galilean Moons, because Galileo was the first to publish a written record of observations of them.
By observing the moons over a period of time, you can figure out the orbital period—the time it takes for the moon to make one complete orbit around Jupiter—of a given moon. Let’s say you do this for Ganymede, and see that it takes about \(P = 7\ \textrm{days}\) to make one complete orbit. By doing more observation (and some math; you also need to know the distance from Earth to Jupiter), you calculate that Ganymede is about \(r = 10^6\ \textrm{km}\) from Jupiter. Using just this information, we can calculate the mass of Jupiter!
The orbit of Ganymede around Jupiter can be approximated as circular. We are dealing with circular motion and the only force acting is gravity, which is pointing towards the center of the circle. Applying Newton’s second law, we have
\[
\begin{align*}
F^G &= ma_c \\
G\frac{m_GM_J}{r^2} &= m_G\frac{v^2}{r} \\
M_J &= \frac{rv^2}{G}
\end{align*}
\]
Now, we need to figure out \(v\). Assuming the speed is constant, the speed is the distance traveled divided by the time it takes to travel that distance—that is, the circumference divided by the period.
\[
\begin{align*}
v &= \frac{C}{P} \\
&= \frac{2\pi r}{P}
\end{align*}
\]
So we have
\[
\begin{align*}
M_J &= \frac{r(2\pi r)^2}{GP^2} \\
&= \frac{2\pi r^3}{GP^2}
\end{align*}
\]
Don’t forget to convert the period from days to seconds, and the distance from kilometers to meters:
\[
\begin{align*}
M_J &= \frac{(2\pi)^2\left(10^9\ \textrm{m}\right)^3}{\left(6.67 \times 10^{-11}\ \textrm{kg}\cdot\textrm{m}^2\textrm{/kg}^2\right)(7 \times 86400\ \textrm{s})^2} \\
&= 1.62 \times 10^{27}\ \textrm{kg}
\end{align*}
\]
NASA tells me that the mass of Jupiter is \(1.89 \times 10^{27}\ \textrm{kg}\). We’re off by about…
\[ \frac{1.89 – 1.62}{1.89} \approx \frac{1.9 – 1.6}{2} = \frac{0.3}{2} = 0.15 \]
…we’re only off by about 15%—that’s really accurate for observational astronomy!