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Satellites, weightlessness and Kepler's laws - TCE Physics (Tasmania)

Syllabus dot point

“Explain apparent weightlessness and apply Kepler's laws to the orbits of satellites and planets.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

Astronauts feel weightless because they and their craft are in free fall together, so there is no normal force, even though gravity still acts. Kepler's laws describe orbits: they are ellipses, sweep equal areas in equal times, and obey T2∝r3T^2 \propto r^3 for bodies around the same central mass.

What this dot point is asking

This dot point applies gravity and circular motion to real orbits, and explains the often misunderstood idea of weightlessness in space.

Apparent weightlessness

Weight is the gravitational force W=mgW = mg, and it does not disappear in orbit. At the altitude of the International Space Station, gravity is still about 9090 per cent of its surface value. What changes is the apparent weight, which is the normal force a surface pushes back with.

In orbit the station and everything in it are in continuous free fall toward Earth, accelerating at the local gg. Because the floor falls with the astronaut, it pushes with zero normal force, and with no normal force there is nothing to sense as weight. This is true weightlessness only in deep space far from any mass; in orbit it is apparent weightlessness caused by free fall.

Orbit is continuous free fall

A satellite is falling toward Earth all the time, but it also moves sideways fast enough that the ground curves away beneath it at the same rate it falls. It keeps missing the Earth, which is what an orbit is.

The same free-fall idea explains the "vomit comet" aircraft used to train astronauts: when the plane flies a parabolic arc and its engines provide only enough thrust to overcome air resistance, everyone inside falls together and feels weightless for around half a minute. No spacecraft is needed; weightlessness is purely a consequence of everything sharing the same gravitational acceleration with no supporting normal force.

Low Earth and geostationary orbits

The orbital speed v=GMrv = \sqrt{\dfrac{GM}{r}} shows that lower orbits are faster. A low Earth orbit has a period of about 9090 minutes. A geostationary orbit is special: its period is exactly one sidereal day, so the satellite stays above a fixed point on the equator, ideal for communications. Setting T=86400 sT = 86400\ \text{s} in Kepler's third law gives a radius of about 4.2×107 m4.2 \times 10^7\ \text{m}, roughly 36 000 km36\,000\ \text{km} above the surface. Low orbits are used for imaging and the space station because they are close and cheap to reach; geostationary orbits are used for broadcast and weather because the satellite appears fixed in the sky.

Kepler's three laws

Kepler described planetary orbits before Newton explained them:

  1. The law of ellipses: each planet moves in an ellipse with the Sun at one focus. For most school problems we approximate orbits as circles.
  2. The law of equal areas: a line from the Sun to a planet sweeps out equal areas in equal times, so a planet moves faster when closer to the Sun.
  3. The law of periods: the square of the orbital period is proportional to the cube of the orbital radius.

The third law follows directly from equating gravity to the centripetal force:

T2=4π2GM r3T^2 = \frac{4\pi^2}{GM}\,r^3

so T2r3\dfrac{T^2}{r^3} is the same constant for every satellite of a given central body, and you can compare two orbits without knowing GG or MM. The second law is a consequence of the conservation of angular momentum, which is why a comet whips quickly past the Sun at perihelion and dawdles at aphelion.

Worked example: comparing two moons

Use Kepler's third law as a ratio

A moon orbits a planet at radius r1r_1 with period T1=2.0T_1 = 2.0 days. A second moon orbits the same planet at four times the radius. Since the constants cancel:

T22T12=r23r13=43=64.\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = 4^3 = 64.

Solve for the period

T2=T164=(2.0)(8)=16 days.T_2 = T_1\sqrt{64} = (2.0)(8) = 16\ \text{days}.

The outer moon takes 1616 days per orbit.

Geostationary orbit

A geostationary orbit is a circular equatorial orbit with a period equal to Earth's rotation period, so the satellite remains above the same point on the surface. Its radius is fixed at about 42 000 km42\,000\ \text{km} from Earth's centre.

Saying there is no gravity in space

A very common misconception is that astronauts float because there is no gravity. Gravity is still strong at orbital heights; weightlessness arises because the astronaut and the spacecraft fall together with the same acceleration, removing the normal force. Always explain it as free fall, not as the absence of gravity.

In the exam, when comparing orbits use Kepler's third law as a ratio so the unknown constants drop out. When asked about weightlessness, state clearly that gravity still acts and that the sensation comes from the absence of a normal force during free fall.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks
The star S2 orbits the invisible object Sgr A* at the centre of our Galaxy. Assume a circular orbit of period 16.116.1 years and radius 1.30×1014 m1.30 \times 10^{14}\ \text{m}. Find the average speed of S2 and the mass of Sgr A*.
Show worked answer →

Average speed is the circumference divided by the period. T=16.1×3.156×107=5.08×108 sT = 16.1 \times 3.156 \times 10^7 = 5.08 \times 10^8\ \text{s}:

v=2πrT=2π(1.30×1014)5.08×108=1.61×106 m s−1 (≈1610 km s−1).v = \frac{2\pi r}{T} = \frac{2\pi (1.30\times10^{14})}{5.08\times10^8} = 1.61 \times 10^6\ \text{m s}^{-1}\ (\approx 1610\ \text{km s}^{-1}).

Mass of Sgr A* from gravity providing the centripetal force, M=v2rGM = \dfrac{v^2 r}{G}:

M=(1.61×106)2(1.30×1014)6.67×10−11=5.05×1036 kg.M = \frac{(1.61\times10^6)^2 (1.30\times10^{14})}{6.67\times10^{-11}} = 5.05 \times 10^{36}\ \text{kg}.

The central object is about 5×1036 kg5 \times 10^{36}\ \text{kg} (roughly 2.52.5 million Suns), consistent with a supermassive black hole. Markers want v=2πrTv = \dfrac{2\pi r}{T} then M=v2rGM = \dfrac{v^2 r}{G}.

Practice3 marks
A planet orbits at radius 9.0×1010 m9.0 \times 10^{10}\ \text{m} from a neutron star of mass 2.79×1030 kg2.79 \times 10^{30}\ \text{kg}. Calculate the orbital period of the planet, in Earth days.
Show worked answer →

Gravity provides the centripetal force, giving Kepler's third law T2=4π2r3GMT^2 = \dfrac{4\pi^2 r^3}{GM}.

T2=4π2(9.0×1010)3(6.67×10−11)(2.79×1030)=2.878×10341.861×1020=1.547×1014.T^2 = \frac{4\pi^2 (9.0\times10^{10})^3}{(6.67\times10^{-11})(2.79\times10^{30})} = \frac{2.878\times10^{34}}{1.861\times10^{20}} = 1.547 \times 10^{14}.

T=1.547×1014=1.244×107 s.T = \sqrt{1.547\times10^{14}} = 1.244 \times 10^7\ \text{s}.

Convert to days: 1.244×10786400=144\dfrac{1.244\times10^7}{86400} = 144 Earth days. Markers want T2=4π2r3GMT^2 = \dfrac{4\pi^2 r^3}{GM} and the conversion of seconds to days.

Practice3 marks
Moon X orbits a planet at radius rr with a period of 2.02.0 days. Moon Y orbits the same planet at four times that radius. Use Kepler's third law to calculate the period of moon Y.
Show worked answer →

Kepler's third law states T2∝r3T^2 \propto r^3 for bodies orbiting the same planet, so use a ratio to cancel the constants:

TY2TX2=rY3rX3=43=64.\frac{T_Y^2}{T_X^2} = \frac{r_Y^3}{r_X^3} = 4^3 = 64.

TY=TX64=(2.0)(8)=16 days.T_Y = T_X \sqrt{64} = (2.0)(8) = 16\ \text{days}.

Moon Y takes 1616 days per orbit. Markers reward using Kepler's third law as a ratio so GG and MM drop out, and the cube-then-square-root arithmetic.

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