Gravitational fields

Every two masses attract each other with a force $F = G\dfrac{m_{1}m_{2}}{r^{2}}$ — proportional to each mass and to the inverse square of the distance r between their centres.

The gravitational **force per unit mass** on a small mass placed in the field: $g = \dfrac{F}{m}$. Unit: **N kg⁻¹**.

$g = G\dfrac{M}{r^{2}}$ — given in the data booklet. M is the source mass, r the distance from its centre.

**N kg⁻¹** — numerically the same as the free-fall acceleration in m s⁻².

Because F = mg and F = ma, so a = g. The falling mass cancels, so g is the acceleration — independent of the mass that falls.

**Inward**, towards the mass — gravity is always **attractive**.

g is divided by **3² = 9** (g is proportional to 1/r², the inverse-square law).

**No** (ignoring air resistance) — the acceleration g = GM/r² doesn't depend on the falling mass, so all masses fall equally fast.

The **gravitational constant**, G = 6.67 × 10⁻¹¹ N m² kg⁻² — the same everywhere in the universe.

About **9.8 N kg⁻¹** (or 9.8 m s⁻²) — found from g = GM/r² using Earth's mass and radius.

g is **inversely proportional to r²** (inverse-square): double r → quarter g; triple r → one-ninth g.

The **square** of a planet's orbital period is **proportional** to the **cube** of its orbital radius: $T^{2} \propto r^{3}$.

Each planet moves in an **ellipse** with the Sun at one **focus** of the ellipse.

A planet moves **faster when nearer the Sun** and **slower when farther away** (it sweeps out equal areas in equal times).

$T^{2} = \dfrac{4\pi^{2}r^{3}}{GM}$ — derived from $g = GM/r^{2}$ and $a = 4\pi^{2}r/T^{2}$. M is the mass being orbited.

Use $\dfrac{T_{A}^{2}}{r_{A}^{3}} = \dfrac{T_{B}^{2}}{r_{B}^{3}}$ — the constant $4\pi^{2}/(GM)$ cancels.

A **straight line through the origin** — because $T^{2}/r^{3}$ is a constant.

$(T_{B}/T_{A})^{2} = 4^{3} = 64$, so $T_{B}/T_{A} = \sqrt{64} = 8$ — **8 times** longer.

By the second law it moves **faster when closer** to the Sun (more kinetic energy) and **slower when farther** (less), so its speed and kinetic energy vary.

The **gravitational pull of the Sun**, directed inward (centripetal), which bends the planet's path into a closed orbit.

T is **squared**, r is **cubed**: $T^{2} \propto r^{3}$. Mixing them up is the classic mistake.

**Gravity** — the inward pull of the central body. There is no separate 'orbit force'.

Pointing **toward the centre** of the circular path. The centripetal force is whatever points inward and curves the motion into a circle.

$\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$ — the orbiting mass m cancels from both sides.

$v = \sqrt{\dfrac{GM}{r}}$ — derived from gravity equalling the centripetal force. A bigger radius gives a smaller speed.

**No** — the mass cancels, so v = √(GM/r) depends only on G, the central mass M and the radius r.

**Toward the central body** (centripetal) — the same direction as gravity.

$T^{2} = \dfrac{4\pi^{2}}{GM}\,r^{3}$ — period squared is proportional to radius cubed; the constant 4π²/GM depends only on the central mass M.

Rearrange Kepler's third law: $M = \dfrac{4\pi^{2} r^{3}}{G\,T^{2}}$ — measure an orbit's period T and radius r.

A circular orbit with a period of exactly **24 h**, in Earth's spin direction, above the **equator** — so the satellite stays above one fixed point.

**height = r − (radius of the planet)**, because r is measured from the planet's centre.

v = √(GM/r) falls as r rises (slower), while T² = (4π²/GM)r³ rises steeply with r (much longer period).

v = √(GM/r) = √[(6.67×10⁻¹¹ × 6.0×10²⁴) / 7.0×10⁶] ≈ 7.6 × 10³ m s⁻¹.

The gravitational potential energy **per kilogram** at a point: $V = -\dfrac{GM}{r}$. Unit: J kg⁻¹. Negative everywhere, zero at infinity.

$E_{p} = -\dfrac{GMm}{r}$ — for a mass m at distance r from a mass M. Unit: joules (J).

We set it to **zero at infinity**; anywhere closer in, gravity has already pulled the object 'downhill', so it has less than zero — it sits in a **well**.

**At infinity** — infinitely far from the mass, where the field has faded to nothing.

The minimum launch speed needed for an object to escape a planet's gravity — to reach where V = 0 (infinitely far) and just stop there.

$v_{esc} = \sqrt{\dfrac{2GM}{r}}$ — from energy conservation. r is usually the planet's radius.

**No** — the mass cancels in v_{esc} = √(2GM/r). It depends only on the planet's mass M and radius r.

Supplying enough **kinetic energy** to climb out of the gravitational well to where V = 0 (infinitely far away): ½mv² = GMm/r.

It **doubles** — v_{esc} ∝ √M, so √4 = 2.

E_{p} becomes **less negative** (rises towards 0), because r increases in E_{p} = -GMm/r.

V is the energy **per kilogram** (J kg⁻¹); E_{p} = mV is the energy of a specific object of mass m (J).

The **work done per unit mass** to bring a small test mass from infinity to a point: $V_g = -\frac{GM}{r}$. Unit: **J kg⁻¹**.

$V_g = -\dfrac{GM}{r}$ — negative, zero at infinity.

$E_p = mV_g = -\dfrac{GMm}{r}$.

The **zero is set at infinity**; a mass **loses** PE moving inward, so everywhere closer is below zero.

At **infinity** — infinitely far from every mass.

$V_g \propto 1/r$ (one over r), while field strength $g \propto 1/r^2$.

A surface joining all points at the **same V_g**; around a point mass these are **concentric spheres**.

Always **90° (perpendicular)**.

**Zero**, because ΔV_g = 0.

$W = m\,\Delta V_g = m(V_{final} - V_{initial})$.

Because **g changes with distance** ($g = GM/r^2$); mgh only works where g is roughly constant.

A **scalar** — add the values from several masses with **no direction**.

$KE = +\dfrac{GMm}{2r}$ — always **positive**.

$PE = -\dfrac{GMm}{r}$ — **negative** (zero at infinity).

$E = -\dfrac{GMm}{2r}$ — the sum KE + PE, and **negative** because the body is bound.

The potential energy is **twice** the size of the kinetic energy: $|PE| = 2\,KE$.

Because the body is **bound** in the gravity well — you'd have to add energy (up to zero) to free it.

$E = -GMm/2r$ becomes **less negative** (rises toward zero) — a higher, more weakly bound state.

The minimum launch speed that lets an object coast to infinity, ending with **zero** total energy.

$v_{esc} = \sqrt{\dfrac{2GM}{r}}$.

**No** — the mass m cancels in ½mv² = GMm/r, so v_esc is the same for any object.

Set launch kinetic energy = depth of the well: ½mv² = GMm/r, then solve for v.

About $1.1\times10^4$ m s⁻¹ ≈ **11 km s⁻¹** (M = 6.0×10²⁴ kg, r = 6.4×10⁶ m).

Escape velocity $\sqrt{2GM/r}$ carries a **factor of 2** under the root; orbital speed $\sqrt{GM/r}$ does not.