Cambridge A-Level Physics 9702 – Gravitational Force Between Point Masses
Gravitational Force Between Point Masses
In Newtonian gravity the attractive force between two point masses \$m1\$ and \$m2\$ separated by a distance \$r\$ is given by
\$\$
F = G\frac{m1 m2}{r^{2}}
\$\$
where \$G = 6.674\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}\$ is the universal gravitational constant.
Uniform Spherical Masses
For many practical problems the object of interest is a solid sphere of uniform density. The key result, proved by the Shell Theorem, is:
Outside a uniform sphere (i.e. at a point where \$r\$ is greater than the sphere’s radius \$R\$), the gravitational field is identical to that produced by a point mass equal to the total mass of the sphere located at its centre.
Inside a uniform spherical shell the net gravitational force is zero; inside a solid sphere the force varies linearly with distance from the centre.
This allows us to treat planets, stars, and laboratory spheres as point masses when we are interested in the field at points external to them.
Derivation Sketch (Shell Theorem)
Consider a thin spherical shell of radius \$R\$ and total mass \$M\$.
Divide the shell into two opposite elements \$d m1\$ and \$d m2\$ that subtend the same solid angle at the external point \$P\$ (distance \$r>R\$ from the centre).
Because the elements are at different distances from \$P\$, their individual forces are not equal, but their components perpendicular to the line joining \$P\$ to the centre cancel.
Summing over the entire shell, only the radial components survive, giving a net force equivalent to that from a point mass \$M\$ at the centre.
Suggested diagram: A point P outside a uniform sphere, showing the sphere’s centre C, radius R, distance r, and two symmetric mass elements on the shell whose perpendicular force components cancel.
Practical Formulae
Quantity
Expression
Applicable Region
Gravitational force (external point)
\$F = G\displaystyle\frac{M m}{r^{2}}\$
\$r \ge R\$
Gravitational field (external point)
\$g = G\displaystyle\frac{M}{r^{2}}\$
\$r \ge R\$
Gravitational force (inside solid sphere)
\$F = G\displaystyle\frac{M m}{R^{3}}\,r\$
\$r \le R\$
Gravitational field (inside solid sphere)
\$g = G\displaystyle\frac{M}{R^{3}}\,r\$
\$r \le R\$
Example Calculation
Find the gravitational force exerted by the Earth (mass \$M{\oplus}=5.97\times10^{24}\,\text{kg}\$, radius \$R{\oplus}=6.37\times10^{6}\,\text{m}\$) on a 70 kg person standing on the surface.
Since the person is on the surface, \$r = R_{\oplus}\$, so we may treat the Earth as a point mass:
This is the familiar weight \$mg\$, confirming the equivalence of the two approaches.
Common Misconceptions
“All mass contributes equally at any distance.” – Only the total mass matters for points outside the sphere; the distribution inside does not affect the external field.
“The force inside a solid sphere is the same as on the surface.” – Inside, the force decreases linearly with distance from the centre because only the mass at radii smaller than the point contributes.
“A hollow sphere behaves like a solid sphere.” – Outside, both give the same external field, but inside a hollow shell the field is zero, unlike a solid sphere.
Practice Questions
A uniform solid sphere of mass \$M=2.0\ \text{kg}\$ and radius \$R=0.10\ \text{m}\$ is placed in space. Calculate the gravitational force on a \$0.5\ \text{kg}\$ particle located \$0.25\ \text{m}\$ from the centre of the sphere.
Show that the gravitational potential energy of a particle of mass \$m\$ at a distance \$r\$ from the centre of a uniform sphere (outside the sphere) can be written as \$U = -G\displaystyle\frac{M m}{r}\$.
A thin spherical shell of mass \$M=5.0\ \text{kg}\$ and radius \$R=0.20\ \text{m}\$ has a small hole through its centre. A \$0.1\ \text{kg}\$ mass is placed at the centre of the hole. What is the net gravitational force on the mass? Explain using the shell theorem.
Derive the expression for the gravitational field inside a uniform solid sphere and verify that it reduces to \$g = G M / R^{2}\$ at the surface.
Summary
For any point external to a uniform sphere, the sphere’s mass can be treated as if it were concentrated at its centre. This simplifies calculations of gravitational force and field, allowing the direct use of Newton’s law for point masses. The result follows from the shell theorem and is fundamental to problems involving planets, stars, and laboratory spheres.