D.1 Gravitational fields

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Newton's law of gravitation

F = G\frac{m_{1}m_{2}}{r^{2}}

Gravitational field strength

g = \frac{F}{m} = G\frac{M}{r^{2}}

Gravitational PE (HL)

E_{p} = -G\frac{m_{1}m_{2}}{r}

Gravitational potential (HL)

V_{g} = -\frac{GM}{r}

Potential gradient (HL)

g = -\frac{\Delta V_{g}}{\Delta r}

Work done (HL)

W = m\Delta V_{g}

Escape speed (HL)

v_{\text{esc}} = \sqrt{\frac{2GM}{r}}

Orbital speed (HL)

v_{\text{orbital}} = \sqrt{\frac{GM}{r}}

Section A — Multiple choice

A1. The gravitational field strength at the surface of a planet is $g$ . At a height equal to the planet's radius above the surface, it is:

A $g/2$
B $g/4$
C $g/8$
D $g/16$

A2. According to Kepler's third law, a planet orbiting at 4 times Earth's orbital radius has a period of:

A 4 years
B 8 years
C 16 years
D 64 years

A3. Gravitational potential is always negative because:

A Gravity always repels
B The zero is defined at infinity and gravity is attractive, so work is released bringing a mass inward
C It is a vector pointing towards the mass
D Potential energy cannot be positive

Section B — Short answer

B1. Using $g = GM/r^2$ at the Earth's surface ( $g = 9.81\ \text{m s}^{-2}$ , $R = 6.37\times10^6\ \text{m}$ ), estimate the mass of the Earth. [3 marks]

B2. Show that the escape speed from a planet is $\sqrt{2}$ times the orbital speed for a circular orbit at its surface. [3 marks]

B3. A satellite in low orbit experiences slight atmospheric drag. Explain why its speed increases even though a resistive force acts on it. [3 marks]

Answer key

For worked explanations, interactive practice and more free resources, visit www.newtonine.com

Section A

A1: $g/4$ — Field strength follows the inverse square law from the planet's CENTRE. At height $R$ the distance from the centre is $2R$ , so $g' = g/2^2 = g/4$ — measure $r$ from the centre, not the surface.

A2: 8 years — $T^2 \propto r^3$ : $T^2 = 4^3 = 64$ , so $T = 8$ years. Cube the radius ratio, then square-root.

A3: The zero is defined at infinity and gravity is attractive, so work is released bringing a mass inward — With $V_g = 0$ at infinite separation and an attractive force doing positive work on an inbound mass, every point closer than infinity sits below zero. The negative sign encodes being bound in a potential well.

Section B

B1: $M = gR^2/G = \frac{9.81 \times (6.37\times10^6)^2}{6.67\times10^{-11}} \approx 6.0\times10^{24}\ \text{kg}$ — Cavendish's "weighing the Earth" in one line.

B2: Escape: total energy zero, $\frac{1}{2}mv_{esc}^2 = GMm/R$ , so $v_{esc} = \sqrt{2GM/R}$ . Orbit: gravity as centripetal force, $mv^2/R = GMm/R^2$ , so $v_{orb} = \sqrt{GM/R}$ . Ratio: $v_{esc}/v_{orb} = \sqrt{2}$ .

B3: Drag removes mechanical energy, so the satellite descends to a lower orbit. But orbital speed $v = \sqrt{GM/r}$ increases as $r$ decreases: the loss in potential energy exceeds the energy removed by drag, and the surplus appears as kinetic energy. Gravity, not drag, does the accelerating.