understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre

Gravitational Force Between Point Masses

Newton’s Law of Gravitation

Newton discovered that every two masses attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula is:

\$F = G \dfrac{m1 m2}{r^2}\$

Where:

• \$F\$ = gravitational force (Newtons, N)
• \$G\$ = gravitational constant (\$6.67\times10^{-11}\,\text{N m}^2\text{/kg}^2\$)
• \$m1, m2\$ = masses (kg)
• \$r\$ = distance between the centres of the masses (m)

Point Masses vs. Uniform Spheres

When we talk about a “point mass”, we imagine all the mass concentrated at a single point. A uniform sphere (like a planet) is more realistic, but for a point that lies outside the sphere, the sphere’s gravity behaves exactly as if all its mass were at its centre. This is a powerful simplification that saves us from complex calculations.

Why Does It Work? (The Shell Theorem)

The shell theorem tells us:

1. A spherically symmetric shell of mass exerts no net gravitational force on a point inside it.
2. A point outside a spherical shell feels the same force as if all the shell’s mass were concentrated at its centre.

Think of the sphere as a collection of many tiny point masses. For a point outside the sphere, the forces from opposite sides cancel out, leaving only the effect of the total mass at the centre. This is like feeling the pull of a giant snowball from far away – you don’t feel the individual snowflakes, just the whole ball’s pull.

Practical Example: Earth and a Satellite

Suppose a satellite of mass \$ms = 500\,\text{kg}\$ orbits Earth at a distance \$r = 7.0\times10^6\,\text{m}\$ from Earth’s centre. Using Earth’s mass \$ME = 5.97\times10^{24}\,\text{kg}\$:

\$F = G \dfrac{ME\,ms}{r^2}\$

Plugging in the numbers gives the gravitational pull on the satellite. Notice we treated Earth as a point mass at its centre – no need to integrate over Earth’s volume!

Quick Check: Is the Point Inside or Outside?

• If the point is outside the sphere (distance > radius), treat the sphere as a point mass at the centre.
• If the point is inside the sphere, the force depends only on the mass enclosed within the radius of the point.

Summary Table

💡 Remember: For any object outside a uniform spherical body, you can treat the entire body as a point mass at its centre. This trick turns a complex problem into a simple one!