Gravitational fields and orbits flashcards

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Gravitational field
A region of space in which a mass experiences a force.
Gravitational field strength, g
The force per unit mass on a small mass placed at a point: g = F/m. A vector; unit N kg⁻¹ (≡ m s⁻²).
Gravitational potential at a point
The work done per unit mass in bringing a small mass from infinity to that point. A scalar; unit J kg⁻¹; zero at infinity.
Equipotential surface
A surface on which every point has the same potential; no work is done moving a mass along it.
Geostationary orbit
A circular orbit above the equator, moving west to east with a period equal to the Earth's rotation period, so the satellite stays above one point on the surface.
Uniform gravitational field
A field with the same strength and direction everywhere: parallel, equally spaced field lines (e.g. close to a planet's surface).
Newton's law of gravitation
F = Gm₁m₂/r², r between centres; G = 6.67 × 10⁻¹¹ N m² kg⁻². (Yes)
Field strength in a radial field
g = Gm/r² (outside a spherical mass, r from its centre). (Yes)
Gravitational potential in a radial field
V = −Gm/r (always negative; zero at infinity). (Yes)
Work done moving a mass m between two points in a field
W = mΔV = m(V_final − V_initial). (No — learn it)
Orbital speed in a circular orbit
v = √(GM/r), from GMm/r² = mv²/r. (No — derive it)
Period and radius of a circular orbit
T² = (4π²/GM) r³, from GMm/r² = mv²/r with v = 2πr/T. (No — derive it)
Mass of a planet from a moon's orbit
M = 4π²r³/(GT²) — T in seconds, r in metres. (No — derive it)
Sketch: g against r outside a planet
Curve falling as 1/r² from the surface value; never reaches zero. At 2R it is one quarter of the surface value.
Sketch: V against r outside a planet
Negative everywhere, rising (less negative) towards zero as r increases, as −1/r. Gradient of the tangent = field strength.
Field lines around a spherical planet
Radial lines pointing in towards the centre, spreading out with distance (field weakens).
Equipotentials around a planet at equal steps of V
Concentric circles, getting further apart with distance; they cross field lines at right angles.
Graph to test T² ∝ r³ for moons of one planet
T² against r³: straight line through the origin, gradient 4π²/GM. Or lg T against lg r: gradient 1.5.
Derive g = GM/r²
Force on test mass m: F = GMm/r². g = F/m, so m cancels: g = GM/r².
Why is gravitational potential negative?
V = 0 at infinity; gravity attracts, so the work done per unit mass bringing a mass in from infinity is negative.
Why do equally stepped equipotentials get further apart away from a planet?
g = −ΔV/Δr: the field weakens with distance, so more distance is needed for the same change in potential.
Gravitational vs electric fields: three similarities
Field strength = force per unit mass/charge; inverse square force and field; potential ∝ 1/r.
Gravitational vs electric fields: three differences
Gravity only attracts (electric forces attract or repel); gravitational potential always negative; gravity far weaker between particles and cannot be cancelled.
Why does gravity dominate on astronomical scales if it is so weak?
Large bodies are electrically neutral, so electric forces cancel; gravity only attracts, so all the mass adds up.
Why does a satellite's mass not affect its orbit?
m cancels in GMm/r² = mv²/r, so v and T depend only on M and r.
Why is there only one geostationary radius?
The period is fixed (one rotation of the Earth), and T² = (4π²/GM)r³ gives one value of r.
Why do astronauts in a low orbit feel weightless?
They and their craft are in free fall together with acceleration g (still ≈ 90% of the surface value), so there is no contact force.
Planet: 2 × Earth's mass, 3 × Earth's radius. Surface g? (g_E = 9.81 N kg⁻¹)
g ∝ M/r²: 9.81 × 2 ÷ 9 = 2.18 N kg⁻¹.
g at 2.0 × 10⁶ m above the Earth (M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m)?
r = 8.37 × 10⁶ m; g = GM/r² = 5.68 N kg⁻¹.
Work to lift 1000 kg from the Earth's surface to r = 2R (M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m)?
W = m(GM/R − GM/2R) = GMm/2R = 3.13 × 10¹⁰ J.
Orbital speed at r = 8.0 × 10⁶ m around the Earth (M = 5.97 × 10²⁴ kg)?
v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ 8.0 × 10⁶) = 7.06 × 10³ m s⁻¹.
Moon B's orbital radius is 9 times moon A's (same planet). Ratio of periods?
T ∝ r^(3/2): 9^1.5 = 27, so B's period is 27 times A's.
A moon orbits at r = 4.0 × 10⁸ m with T = 2.0 × 10⁵ s. Mass of the planet?
M = 4π²r³/(GT²) = 9.5 × 10²⁶ kg.
Satellite at height h above a planet of radius R: what is r?
r = R + h. Every equation here uses the distance from the centre.
Distance multiplied by k: what happens to F, g, V, v and T?
F and g ÷ k²; V ÷ k; orbital speed ÷ √k; period × k^(3/2).
When can you use mgΔh instead of mΔV?
Only when the height change is small compared with the planet's radius, so g is effectively constant.
“Derive T² ∝ r³” — the first line
State that the gravitational force provides the centripetal force, then write GMm/r² = mv²/r before substituting v = 2πr/T.
Period given in days or hours
Convert to seconds before using G: 1 day = 8.64 × 10⁴ s.

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