Know that the strength of the gravitational field (a) at the surface of a planet depends on the mass of the planet (b) around a planet decreases as the distance from the planet increases

6.1.2 The Solar System

Learning Objective

Understand how the strength of the gravitational field (a) at the surface of a planet depends on the planet’s mass and (b) decreases with increasing distance from the planet.

Key Concepts

• Gravitational field strength \$g\$ is the force per unit mass: \$g = \dfrac{F}{m}\$.
• For a spherical planet of mass \$M\$ and radius \$R\$, the field at the surface is \$g = \dfrac{GM}{R^{2}}\$, where \$G = 6.67\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}\$.
• At a distance \$r\$ from the centre of the planet, the field is \$g(r) = \dfrac{GM}{r^{2}}\$ (inverse‑square law).

Why Mass Affects Surface Gravity

The larger the mass \$M\$, the greater the numerator in \$g = GM/R^{2}\$, giving a stronger field, provided the radius does not increase proportionally.

Example: Earth vs. Mars

How Gravitational Field Decreases with Distance

From the inverse‑square law, if the distance from the planet’s centre is doubled, the field strength becomes one‑quarter of its original value.

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g(2r) = \frac{GM}{(2r)^{2}} = \frac{1}{4}\,\frac{GM}{r^{2}} = \frac{1}{4}\,g(r)

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1. Choose a reference distance \$r_{0}\$ (usually the planet’s radius).
2. Calculate \$g(r)\$ for any \$r\$ using \$g(r)=g(r{0})\left(\dfrac{r{0}}{r}\right)^{2}\$.

Worked Example

Find the gravitational field strength at a height of 400 km above Earth’s surface.

Given:

• Earth’s surface \$g_{0}=9.81\ \text{m s}^{-2}\$
• Earth’s radius \$R=6.37\times10^{6}\ \text{m}\$
• Height \$h=4.0\times10^{5}\ \text{m}\$

Distance from centre \$r = R + h = 6.77\times10^{6}\ \text{m}\$.

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g(r) = g_{0}\left(\frac{R}{R+h}\right)^{2}

= 9.81\left(\frac{6.37\times10^{6}}{6.77\times10^{6}}\right)^{2}

\approx 8.7\ \text{m s}^{-2}

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Summary Checklist

• Surface gravity \$g\$ increases with planetary mass and decreases with the square of the radius.
• At any point outside a spherical planet, \$g\$ follows the inverse‑square law \$g\propto 1/r^{2}\$.
• Doubling the distance reduces \$g\$ to one‑quarter; tripling reduces it to one‑ninth, etc.