Gravitational Fields

Overview

Gravitation describes the universal attractive interaction between masses. It explains:

• why objects fall toward Earth,
• why planets orbit the Sun,
• why moons orbit planets,
• why satellites remain in orbit.

Gravitational interactions are long-range and act through a gravitational field. This topic connects ideas from Forces, Dynamics, Circular Motion, and Work, Energy and Power.

Core Ideas

Newton’s Law of Gravitation

Any two point masses attract each other. For masses $m_1$ and $m_2$ separated by centre-to-centre distance $r$:

$$F=\frac{G{m}_1{m}_2}{r^2}$$

where

$$G=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-2}$$

If $\hat{r}$ is a unit vector directed outward from source mass $M$ to test mass $m$, then the force on $m$ is:

$${\vec{F}}_G=-\frac{GMm}{r^2}\hat{r}$$

The negative sign shows the force acts toward $M$.

Figure: The gravitational force on test mass $m$ is directed toward source mass $M$.

Key Features

• Always attractive.
• Acts along the line joining centres.
• Obeys inverse-square law:

$$F\propto \frac{1}{r^2}$$

• If $r$ doubles, force becomes $\frac{1}{4}$.
• Applies exactly to point masses and spherically symmetric bodies.

Gravitational Field Strength

A gravitational field is a region where a mass experiences gravitational force.

Gravitational field strength is force per unit mass:

$$\vec{g}=\frac{{\vec{F}}_G}{m}$$

For a point mass $M$:

$$\vec{g}=-\frac{GM}{r^2}\hat{r}$$

Magnitude:

$$g=\frac{GM}{r^2}$$

Units:

• $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}$
• $\mathrm{m}{\mathrm{s}}^{-2}$

Near Earth’s Surface

Near Earth’s surface:

• height changes are small compared with Earth’s radius,
• field lines are approximately parallel,
• $g$ is nearly constant.

Hence:

$$g\approx 9.81\mathrm{m}{\mathrm{s}}^{-2}$$

Often taken as:

$$g=9.8\mathrm{m}{\mathrm{s}}^{-2}$$

Superposition of Fields

Gravitational field strength is a vector quantity.

If several masses are present:

$${\vec{g}}_{\text{net}}={\vec{g}}_1+{\vec{g}}_2+{\vec{g}}_3+\cdots$$

Thus:

• direction matters,
• fields may reinforce,
• fields may cancel at certain points.

Gravitational Potential

Gravitational potential $V$ at a point is the work done per unit mass by an external agent in bringing a small test mass from infinity to that point without change in kinetic energy.

For a point mass:

$$V=-\frac{GM}{r}$$

Units:

$$\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$$

Key Ideas

• Potential is a scalar.
• Negative when zero is taken at infinity.
• Becomes less negative further away.

Superposition

$$V_{\text{net}}=V_1+V_2+V_3+\cdots$$

Potentials add algebraically.

Gravitational Potential Energy

For a mass $m$ placed at potential $V$:

$$U=mV$$

Hence:

$$U=-\frac{GMm}{r}$$

Meaning

• Energy of the mass-position system.
• Scalar quantity.
• Negative for bound systems.

Energy must be supplied to separate masses to infinity.

Relation Between Field and Potential

Force and field:

$${\vec{F}}_G=m\vec{g}$$

Field and potential:

$$\vec{g}=-\nabla V$$

For radial motion:

$$g=-\frac{dV}{dr}$$

Force and potential energy:

$${\vec{F}}_G=-\nabla U$$

For one-dimensional radial motion:

$$F_r=-\frac{dU}{dr}$$

Circular Orbits

For a satellite of mass $m$ orbiting mass $M$ at radius $r$:

$$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$$

So:

$$v=\sqrt{\frac{GM}{r}}$$

Angular speed:

$$\omega =\frac{v}{r}=\sqrt{\frac{GM}{r^3}}$$

Orbital period:

$$T=\frac{2\pi r}{v}$$

Hence:

$$T^2=\frac{4{\pi }^2}{GM}{r}^3$$

Figure: In circular orbit, ${\vec{F}}_G$ is radial and $\vec{v}$ is tangential.

Higher orbit:

• lower speed,
• longer period.

See Orbital Motion in Gravity.

Satellite Energy

For circular orbit:

$$K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$$ $$U=-\frac{GMm}{r}$$

Total energy:

$$E=K+U=-\frac{GMm}{2r}$$

Since $E<0$, the satellite is gravitationally bound.

If orbital energy decreases (e.g. drag):

• radius decreases,
• speed increases,
• satellite spirals inward.

Escape Velocity

Minimum launch speed from distance $R$ from centre to reach infinity with zero final speed.

Using conservation of energy:

$$\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0$$

Thus:

$$v_e=\sqrt{\frac{2GM}{R}}$$

For Earth:

$$v_e\approx 11.2\mathrm{km}{\mathrm{s}}^{-1}$$

See Escape Velocity.

Geostationary Orbit

A geostationary satellite remains above the same point on Earth.

Conditions:

• circular orbit,
• above equator,
• west to east,
• period $\approx 24\mathrm{h}$.

Uses:

• communications,
• weather monitoring,
• broadcasting.

See Geostationary Orbit.

Worked Examples

Example 1: Field Strength at Earth’s Surface

Using:

$$g=\frac{GM}{R^2}$$

For Earth:

• $M=5.97\times 10^{24}\mathrm{kg}$
• $R=6.37\times 10^6\mathrm{m}$

Gives:

$$g\approx 9.8\mathrm{m}{\mathrm{s}}^{-2}$$

Example 2: Orbital Speed

A satellite orbits Earth at radius $r$.

$$v=\sqrt{\frac{GM}{r}}$$

So increasing $r$ decreases orbital speed.

Example 3: Escape Speed from Moon

Use:

$$v_e=\sqrt{\frac{2GM}{R}}$$

Same method as Earth, but Moon gives much smaller value.

Exam Relevance

Common Exam Pitfalls

• Confusing $g$ with $G$.
• Confusing $\vec{g}$ (vector) with $V$ (scalar).
• Using positive gravitational potential.
• Forgetting $r=R+h$.
• Thinking gravity is zero in orbit.
• Thinking higher orbit means higher speed.
• Using constant $g$ far from Earth.
• Confusing potential with potential energy.

Formula Summary

$$F=\frac{G{m}_1{m}_2}{r^2}$$ $$\vec{g}=-\frac{GM}{r^2}\hat{r}$$ $$V=-\frac{GM}{r}$$ $$U=-\frac{GMm}{r}$$ $$v=\sqrt{\frac{GM}{r}}$$ $$T^2=\frac{4{\pi }^2}{GM}{r}^3$$ $$v_e=\sqrt{\frac{2GM}{R}}$$

Links

Related Links

• Orbital Motion in Gravity
• Escape Velocity
• Gravitational vs Electric Fields
• Circular Motion
• Work, Energy and Power
• Electric Fields