gravitation Kepler laws and cosmic velocities

Newton showed that two point masses attract each other along the line joining them with the force $F = G\,\frac{m_1 m_2}{r^{2}},\qquad G \approx 6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}.$

For a body on a circular orbit of radius $r$ this force plays the role of the centripetal force, so $G M m / r^{2} = m v^{2}/r$, hence

$$v = \sqrt{\frac{GM}{r}}.$$

Newton's law of gravitation applied to general (not necessarily circular) two-body motion together with Newton's second law yields Kepler's three laws: a planet moves along an ellipse with the Sun at one focus; the radius vector sweeps equal areas in equal times; the squares of the periods are proportional to the cubes of the semi-major axes, $T^{2}/a^{3} = \mathrm{const}$. The circular-orbit relation above is a special case: it reproduces the third law only for circular motion ($T^{2} \propto r^{3}$), while the first and second laws require non-circular motion.

Kepler's second law is conservation of angular momentum: at perihelion and aphelion $v\,r = \mathrm{const}$, so a comet moves faster when it is close to the Sun.

Equating gravitation to the "weight" at a planet's surface gives the first cosmic velocity: $v_1 = \sqrt{gR} = \sqrt{GM/R} \approx 7.9\ \mathrm{km/s}$ for the Earth.

The second cosmic velocity follows from the zero-total-energy condition $\frac{1}{2}v^{2} - GM/R = 0$, hence $v_2 = \sqrt{2GM/R} = \sqrt{2}\,v_1 \approx 11.2\ \mathrm{km/s}$.

A satellite of mass $m = 500$ kg moves in a circular orbit of radius $r = 1.28\times10^{7}$ m around the Earth. Taking $GM = 3.986\times10^{14}\ \mathrm{m^3/s^2}$, find the gravitational force acting on the satellite and the gravitational field strength (free-fall acceleration) at that height.

Find the radius of the geostationary orbit, on which a satellite circles the Earth with period $T = 8.64\times10^{4}$ s (24 h), and its speed. Use Kepler's third law and $GM = 3.986\times10^{14}\ \mathrm{m^3/s^2}$.

A comet moves around the Sun along an ellipse with perihelion distance $r_p = 1.0\times10^{11}$ m and aphelion distance $r_a = 5.0\times10^{11}$ m. Using Kepler's second law and conservation of energy, find the comet's speed at perihelion and at aphelion. Take $GM_{\odot} = 1.33\times10^{20}\ \mathrm{m^3/s^2}$.

Self-study work in a simulation (the safe route: real orbital measurements are out of reach).

1. Open the PhET simulation "Gravity and Orbits" or an equivalent one.
2. Keep the central mass fixed and create at least five circular orbits with different radii $r$; record $r$ and the period $T$ for each.
3. For each orbit compute $r^{3}$ and $T^{2}$ and plot $T^{2}$ against $r^{3}$.
4. Check that the points lie on a straight line through the origin and find the slope $k = T^{2}/r^{3}$; compare it with $4\pi^{2}/(GM)$.
5. Halve the central mass, repeat one measurement and verify that the slope doubles — this is Kepler's third law.

Show the geometric meaning of Kepler's laws and that the third law $T^{2} \propto a^{3}$ is verified by a straight line on a graph.

A two-panel diagram.

• Left — an elliptical orbit with the Sun at one focus; two sectors swept by the radius vector in equal time intervals are shaded and have equal areas (the second law).
• Right — a graph: $a^{3}$ on the horizontal axis, $T^{2}$ on the vertical axis; the points for several planets lie on a straight line through the origin (the third law).

The International Space Station flies about $400$ km above the Earth's surface. With the Earth's radius $R \approx 6400$ km the orbital radius is $r \approx 6800$ km $= 6{,}8\cdot10^{6}$ m, and $GM = 3{,}99\cdot10^{14}$ m$^3$/s$^2$. Then the station's speed is

$$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{3{,}99\cdot10^{14}}{6{,}8\cdot10^{6}}}\approx 7{,}7\ \text{km/s},$$

and the period is $T = 2\pi r/v \approx 5550$ s $\approx 92$ min. In one day the crew sees about 16 sunrises.

This is the first cosmic (circular) velocity at work: the gravitational force $F=GMm/r^2$ plays the role of the centripetal force, giving $v=\sqrt{GM/r}$. It is a direct consequence of Newton's second law applied to circular motion.

A satellite-TV dish points at a satellite that hangs in one fixed spot in the sky. Its period must equal the sidereal day: $T = 86\,164$ s. From Kepler's third law

$$a=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}}}=\sqrt[3]{\frac{3{,}99\cdot10^{14}\cdot(86\,164)^{2}}{4\pi^{2}}}\approx 4{,}22\cdot10^{7}\ \text{m},$$

that is $42\,200$ km from the Earth's centre and an altitude of $\approx 35\,800$ km. The satellite's speed is $v = 2\pi a/T \approx 3{,}1$ km/s.

The example tests Kepler's third law $T^{2} = 4\pi^{2}a^{3}/(GM)$: only at one radius does the orbital period equal one day, so the geostationary orbit is unique. The factor $GM$ is the same gravitational constant as in the law of universal gravitation.

A launch vehicle accelerates its stage to $7{,}9$ km/s to reach low orbit, and to $11{,}2$ km/s to leave the Earth for good. For a flight to Mars the engine on the low orbit ($h=200$ km) is fired again, adding $\approx 3{,}6$ km/s. The trip follows half of a Hohmann ellipse: the Earth at $1{,}00$ AU, Mars at $1{,}52$ AU, so the semi-major axis is $a = (1{,}00+1{,}52)/2 = 1{,}26$ AU. By Kepler's third law $T = a^{3/2} = 1{,}42$ years, so the transfer takes $T/2 \approx 0{,}71$ years $\approx 259$ days.

The second cosmic velocity $v_{2}=\sqrt{2GM/R}=11{,}2$ km/s follows from energy conservation: the kinetic energy exactly cancels the gravitational potential energy $-GMm/R$. The shape of the trajectory and the flight time are set by Kepler's first and third laws; near perihelion (at the Earth) the craft moves faster than near aphelion — that is Kepler's second law.