Flashcards · 9702 · A Level Topic 13

Gravitational fields (A Level) flashcards

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Gravitational field: A region of space in which a mass experiences a force due to another mass. A field of force.
What do gravitational field lines show?: The direction of the force on a small test mass. They are drawn closer together where the field is stronger.
Field-line pattern around an isolated spherical mass: Radial lines pointing in towards the centre of the mass, getting further apart with distance (the field weakens).
Field-line pattern close to the Earth’s surface: Parallel, equally spaced lines pointing vertically down: the field is approximately uniform over small height changes.

Newton’s law of gravitation: F = Gm₁m₂ ÷ r², an attractive force between two point masses, where r is the distance between their centres. G = 6.67 × 10⁻¹¹ N m² kg⁻².
The separation of two masses doubles. What happens to the gravitational force?: It falls to a quarter: an inverse square law, F ∝ 1/r².
Derive the orbital speed of a satellite in a circular orbit: Gravity provides the centripetal force: GMm ÷ r² = mv² ÷ r, so v = √(GM ÷ r). The satellite’s mass m cancels.
Derive T² ∝ r³ for a circular orbit: v² = GM ÷ r and v = 2πr ÷ T, so 4π²r² ÷ T² = GM ÷ r, giving T² = (4π² ÷ GM) r³.
Three conditions for a geostationary orbit: Period of 24 hours; orbiting west to east (the same direction as the Earth spins); directly above the Equator. Radius about 42 000 km from the Earth’s centre.

Gravitational field strength: The force per unit mass on a small test mass placed at that point: g = F ÷ m. Unit N kg⁻¹. A vector.
Field strength of a point mass M: g = GM ÷ r², with r measured from the centre of the mass. For a satellite at height h above a planet of radius R, r = R + h.
Why can g = GM/r² be used for the field at the Earth’s surface?: Outside a uniform sphere, the mass acts as if concentrated at its centre. Near the surface r ≈ R, so g is roughly constant.
Earth: M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m. Find g at the surface.: g = GM ÷ R² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (6.37 × 10⁶)² = 9.8 N kg⁻¹.
A planet has mass 6.4 × 10²³ kg and radius 3.4 × 10⁶ m. Find g at its surface.: g = (6.67 × 10⁻¹¹ × 6.4 × 10²³) ÷ (3.4 × 10⁶)² = 3.7 N kg⁻¹.

Gravitational potential: The work done per unit mass in bringing a small test mass from infinity to that point. φ = −GM ÷ r. Unit J kg⁻¹. A scalar.
Why is gravitational potential always negative?: It is defined as zero at infinity, and gravity is attractive: work must be done on a mass to move it out to infinity, so at any finite distance the potential is below zero.
Gravitational potential energy of a mass m at distance r from M: Ep = mφ = −GMm ÷ r.
How does gravitational potential change as r increases?: It increases (becomes less negative), approaching zero at infinity.
Field strength vs potential for a point mass: g = GM/r²: a vector, falls off as 1/r². φ = −GM/r: a scalar, falls off as 1/r. g = −dφ/dr: field strength is minus the potential gradient.