# 7.2.2 Gravitational Field of a Point Mass

Consider a massive astronomical body of mass M and a tiny satellite of mass m, separated by distance r.

We can say that M is exerting a gravitational force of $\displaystyle {{F}_{g}}=\frac{{GM}}{{{{r}^{2}}}}m$ on m, as according to Newton’s Law of Universal Gravitation.

Alternatively, we can say that M is creating a gravitational field g around it, causing m to experience a gravitational force of mg.

Since gravitational field strength is force per unit mass, the strength of the gravitational field produced by M is

\displaystyle \begin{aligned}g&=\frac{{GMm}}{{{{r}^{2}}}}\div m\\&=\frac{{GM}}{{{{r}^{2}}}}\end{aligned}

A field can be represented by field lines, where the arrows indicate the direction of the field, and the density of field lines (number of lines per unit area) indicate the field strength. The gravitational field produced by M can thus be represented by the field pattern shown below. It is not a uniform one. Direction-wise, it is an attractive field centred about M, directed radially inward and towards M’s centre. Magnitude-wise, it decreases with distance from M. Just like gravitational force, gravitational field strength obeys the inverse-square law (which is a more stylo way of saying something is inversely proportional to the square of the distance).

Two things worth nothing:

1. You may see the formula written with a negative sign, i.e. $\displaystyle g=-\frac{{GM}}{{{{r}^{2}}}}$. The negative sign is supposed to indicate that g is directed inward (in opposite direction to r). But the negative sign is really unnecessary. When we see $\displaystyle g=\frac{{GM}}{{{{r}^{2}}}}$, we understand that we are merely writing down the magnitude of g.
2. We derived this formula assuming that M is a point mass (the assumption is implicit when we use the formula $\displaystyle {{F}_{g}}=\frac{{GMm}}{{{{r}^{2}}}}$). Massive astronomical bodies (stars, planets, etc) are of course not point masses. But they are spherical in shape and have a spherical symmetry, and can be shown mathematically to be equivalent to a point mass situated at the centre of the sphere to someone outside the sphere. So this formula is valid for stars and planets as well, but for points outside the spherical surface only.

Our very own Earth can be modelled as a uniform sphere of radius  ${{R}_{E}}=6370\text{ km}$ and mass ${{M}_{E}}=5.97\times {{10}^{{24}}}\text{ kg}$. The theoretical values of $\displaystyle g=\frac{{G{{M}_{E}}}}{{{{r}^{2}}}}$ for $r\ge {{R}_{E}}$ can thus be calculated.

The graph above highlights the fact that gravitational field strength obeys the inverse-square law. The value of g at $\displaystyle r=2{{R}_{E}}$ is $\frac{1}{4}$ the value of g at $r={{R}_{E}}$. On the other hand, it also highlights the fact that value of g on Earth is practically constant. Whether on top of Everest (~+8 km) or at the bottom of the Pacific Ocean (~−10 km), the percentage variation in the value of r (~6370 km) is not significant.

Explanation Video

What is a Gravitational Field?

Concept Test

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