Universal gravitation and orbital motion

Any two point masses attract each other along the line joining them with a force $F = G\dfrac{m_1 m_2}{r^2}$, where $G = 6.674\times10^{-11}\ \text{N·m}^2/\text{kg}^2$ and $r$ is the distance between the centres of the bodies.

If there are several bodies, the net force on each is found by vector addition of the individual forces, following the superposition principle, just as in applying Newton's second law.

For a satellite in a circular orbit, gravity is the only force and it also plays the role of the centripetal force: $G\dfrac{Mm}{r^2} = m\dfrac{v^2}{r}.$

The satellite mass $m$ cancels, so the speed and the period depend only on the mass of the central body $M$ and the orbital radius $r$ measured from the centre, not from the surface: $v = \sqrt{\dfrac{GM}{r}}, \qquad T = 2\pi\sqrt{\dfrac{r^3}{GM}}.$

This gives Kepler's third law: for all satellites of one central body, $T^2 \propto r^3$.

The Earth has mass $M = 5.97\times10^{24}$ kg. Find the gravitational force between the Earth and a satellite of mass $m = 1000$ kg located at a distance $r = 7.0\times10^{6}$ m from the Earth's centre.

A satellite moves in a circular orbit of radius $r = 7.0\times10^{6}$ m around the Earth ($M = 5.97\times10^{24}$ kg). Find the satellite's speed and its orbital period.

A geostationary satellite must have a period $T = 86164$ s (one sidereal day) and move in the equatorial plane. Determine its orbital radius and its altitude above the Earth's surface. Given: $M = 5.97\times10^{24}$ kg, Earth radius $R_E = 6.37\times10^{6}$ m.

Self-check with a simulation. Real orbital launches are unavailable and hazardous for independent study, so use a model: the PhET simulation "Gravity and Orbits" in "Model" mode.

1. Turn on the gravity force and velocity vectors. Confirm that the force points toward the central body and the velocity is tangent to the orbit.
2. With the central body's mass fixed, record the circular orbital radius $r$ and the measured period $T$ for 4–5 different radii.
3. Plot $T^2$ against $r^3$. The points should fall on a straight line through the origin.
4. Estimate the slope of the line and compare it with the theoretical value $4\pi^2/(GM)$.
5. Double the mass of the central body and verify that at the same radius the period decreases by a factor of $\sqrt{2}$.

Show that gravity acts as the centripetal force in a circular orbit, and how the speed and period depend on the orbital radius.

Diagram and graph.

• Diagram: a massive body $M$ at the centre; a satellite $m$ at distance $r$ from the centre. From the satellite toward the centre, a force vector $\vec{F}$ of magnitude $G\dfrac{Mm}{r^2}$; along the orbit, a velocity vector $\vec{v}$ perpendicular to $\vec{F}$. Dashed lines show two orbits of different radii labelled $r_1 < r_2$.
• Graph: $r^3$ on the horizontal axis, $T^2$ on the vertical axis; plotted data points and a straight line through the origin with slope $4\pi^2/(GM)$. Caption: "$T^2 \propto r^3$ — Kepler's third law".

The International Space Station flies about $400$ km above Earth's surface. Earth's radius is $R = 6{,}371\times10^{6}$ m, so the distance from Earth's centre to the station is $r = 6{,}77\times10^{6}$ m. Earth's mass is $M = 5{,}97\times10^{24}$ kg and $G = 6{,}67\times10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. The calculation gives a speed $v=\sqrt{GM/r}\approx 7{,}7$ km/s and a period $T=2\pi r/v\approx 5{,}5\times10^{3}$ s, that is about $92$ minutes.

The central part of the theory is at work here: the only force on the station is Earth's pull, $F=GMm/r^2$, and it plays the role of the centripetal force $F=mv^2/r$. Setting them equal gives $v=\sqrt{GM/r}$: the station's mass $m$ cancels, so at a given altitude every body moves with the same speed.

Your phone fixes its position from signals of GPS satellites flying at an altitude of $20\,200$ km, that is at a distance $r = 2{,}66\times10^{7}$ m from Earth's centre. With $M = 5{,}97\times10^{24}$ kg we get $v=\sqrt{GM/r}\approx 3{,}9$ km/s, and the period $T=2\pi r/v\approx 4{,}3\times10^{4}$ s $\approx 12$ h. The satellite makes two revolutions per day, so it is above the same point of Earth at the same time each day.

The example tests how orbital quantities depend on radius: comparing the GPS satellite with the ISS, we see that when $r$ grows by a factor of $3{,}9$, the speed falls by about a factor of $2$ ($7{,}7/3{,}9\approx 2$) while the period grows. This is Kepler's third law $T^2\propto r^3$, which follows directly from $F=GMm/r^2$ and $F=mv^2/r$.

A high jumper of mass $70$ kg takes off at $5{,}0$ m/s and rises $1{,}3$ m. At Earth's surface $g=GM/R^2=6{,}67\times10^{-11}\cdot5{,}97\times10^{24}/(6{,}371\times10^{6})^2\approx 9{,}8$ m/s$^2$, so their weight is $mg\approx 687$ N. During the jump the distance to Earth's centre changes by $1{,}3$ m, that is by the fraction $1{,}3/6{,}371\times10^{6}\approx 2\times10^{-7}$; by the inverse-square law the weight changes by only $70\cdot9{,}8\cdot4\times10^{-7}\approx 3\times10^{-4}$ N.

The example shows that "weight" is the same universal-gravitation force $F=GMm/R^2$, only applied at a planet's surface, and that the free-fall acceleration $g=GM/R^2$ does not depend on the jumper's mass. The $1/R^2$ proportionality explains why the change in $g$ can be neglected at jump height but not at ISS altitude.