DPST1021 Chap.8 Gravitation, Satellites and Orbits
Gravitation, Satellites and Orbits
Gravitation closes the mechanics block of Physics 1A. Two lectures in week 6 take Newton’s inverse-square law from a single pair of masses to planets, moons and satellites, and Test 1 draws on them alongside everything from the first seventeen lectures.
Students meet the universal constant G, learn why the field outside a uniform sphere acts as though the whole mass sits at its centre, and see how the acceleration of free fall depends on distance from that centre.
Variation in g gets particular attention.
Altitude weakens it, the spin of the Earth reduces the apparent value at the equator, and an uneven or flattened planet adds further differences, so the familiar 9.80 m/s squared on the data sheet is a convenient average rather than a constant of nature. Superposition then extends the force law to several bodies at once: each attraction is found separately and the vectors are added.
Energy is the second theme.
Far from the surface, mgh no longer describes potential energy, and the formula sheet replaces it with a negative expression that reaches zero only at infinite separation. Combining that expression with the centripetal condition for a circular orbit gives neat relationships between kinetic, potential and total energy, and these explain why moving a satellite to a higher orbit costs energy even though it ends up slower.
Practice in these pages uses the Earth and Moon values printed on the course data sheet.
What this chapter covers
- 01
Newton’s law of gravitation: calculate the attractive force between two masses from G, the product of the masses and the inverse square of their separation.
- 02
Field strength and altitude: express g as GM/r squared outside a sphere, with r measured from the centre, and predict how it changes with height.
- 03
Rotation and shape effects: account for the centripetal share of gravity at the equator and explain why measured g differs between locations on Earth.
- 04
Superposition of attractions: add forces from two or more bodies as vectors, and locate points where competing attractions cancel along a line.
- 05
Potential energy on large scales: apply U equals minus Gm1m2 over r, take differences between two radii, and recognise when mgh is an acceptable shortcut.
- 06
Energy of circular orbits: derive orbital speed from the centripetal condition and relate kinetic, potential and total energy at any orbital radius.
Worked example · free
Speed, period and energy of a low orbit
- 1Orbit radius from the centre: r = 6.371 × 10⁶ + 2.000 × 10⁶ = 8.371 × 10⁶ m.
- 1Gravity provides the centripetal force, so v² = GM/r = (6.674 × 10⁻¹¹)(5.972 × 10²⁴) / 8.371 × 10⁶ = 4.761 × 10⁷ m²/s², giving v = 6.900 × 10³ m/s.
- 1Period: T = 2πr / v = 2π(8.371 × 10⁶) / 6900 = 7622 s, or 127.0 minutes.
- 1Kinetic energy: K = ½mv² = 0.5 × 350 × 4.761 × 10⁷ = 8.33 × 10⁹ J.
- 1Potential energy U = −GMm/r = −1.67 × 10¹⁰ J, exactly −2K, so the total is E = K + U = −8.33 × 10⁹ J.
Key terms
- Gravitational constant
- The universal constant G = 6.674 × 10⁻¹¹ N m²/kg² that sets the strength of the attraction between any two masses.
- Gravitational field strength
- The force per kilogram that a mass experiences at a point, equal to GM/r² outside a spherical body and measured in m/s² or N/kg.
- Superposition principle
- The rule that the net gravitational force on a body is the vector sum of the separate forces exerted by every other mass.
- Gravitational potential energy
- The energy of a pair of masses due to their separation, given by −Gm₁m₂/r and taken as zero when they are infinitely far apart.
- Orbital period
- The time for one complete revolution of a satellite, equal to the circumference 2πr divided by the orbital speed.
Gravitation, Satellites and Orbits FAQ
Why is gravitational potential energy negative?
Zero is chosen at infinite separation, and gravity is attractive, so bringing masses closer releases energy and leaves U below zero. Only differences matter in a calculation. A satellite with negative total energy is bound and would need extra energy to reach zero before escaping.
Does a heavier satellite need a different speed for the same orbit?
No. The satellite’s mass appears on both sides of the centripetal condition and cancels, so speed and period depend only on the planet’s mass and the orbit radius. A heavier satellite does carry proportionally more energy of every kind, so launching it costs more fuel.
When can I still use mgh for gravitational potential energy?
Use it while the height change is tiny compared with the Earth’s radius, so that g barely varies. A 1 km climb changes g by about 0.03%, which is negligible. Once heights reach hundreds of kilometres, g has fallen by several percent and the full expression from the formula sheet is required.
Why do astronauts float if gravity still acts on them in orbit?
At typical space-station altitudes gravity is still close to 90% of its surface value. Astronauts float because they and their spacecraft are falling together, both accelerating toward the Earth at the same rate while moving sideways fast enough to keep missing it. Nothing presses them against the floor.
Exam move
Begin with the inverse-square law in ratio form: if r doubles, the force drops to a quarter,, a scaling that checks answers quickly. Next, make a habit of converting every altitude into a distance from the centre before substituting. For potential energy, set up each problem as a difference between two radii and compare the result with mgh to see when the shortcut works.
Derive the orbit relations yourself from one equation, gravitational force equals mass times centripetal acceleration, , because the formula sheet prints only the force and potential energy. Sketch the energy curves for K, U and E against r once from memory. Quiz 5 covers these two lectures, so repeat its gravitation questions until the steps feel automatic, and carry four significant figures until the last line.
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