recall and use g = GM / r

Gravitational Force Between Point Masses

Learning Objective

Recall and apply the relationship

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

where g is the gravitational field strength at a distance r from a mass M.

Key Concepts

• Newton’s law of universal gravitation
• Definition of gravitational field strength (g)
• Using g = GM/r to find acceleration due to gravity at any point outside a spherical mass
• Assumptions: point masses or spherically symmetric bodies

Newton’s Law of Universal Gravitation

The attractive force between two point masses M and m separated by a distance r is

\$F = \frac{GMm}{r^{2}}\$

where G is the universal gravitational constant, G = 6.674 \times 10^{-11}\,\text{N·m}^{2}\text{kg}^{-2}.

Gravitational Field Strength (g)

The gravitational field strength at a point is the force per unit test mass placed at that point:

\$g = \frac{F}{m} = \frac{GM}{r^{2}} \times \frac{1}{m} \times m = \frac{GM}{r^{2}}\$

For a point on the surface of a spherical body (or at a distance r from its centre) the field strength simplifies to

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

Multiplying both sides by r gives the useful form

\$g = \frac{GM}{r} \times \frac{1}{r} = \frac{GM}{r^{2}}\$

When the distance r is measured from the centre of the body, the expression g = GM/r is often used in the context of orbital motion, where the centripetal acceleration required for a circular orbit is v^{2}/r = GM/r^{2}. Rearranging gives v^{2}=GM/r, and the orbital speed can be expressed as v = \sqrt{GM/r}.

Derivation of g = GM / r for Orbital Motion

1. For a body of mass m moving in a circular orbit of radius r around a much larger mass M, the required centripetal force is F_c = mv^{2}/r.
2. Set the gravitational force equal to the centripetal force:

   \$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}\$

3. Cancel m and one factor of r:

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

4. Recognise that the orbital speed squared, v^{2}, is the product of the gravitational field strength g and the radius r:

   \$v^{2}=gr\$

5. Substituting gives:

   \$gr = \frac{GM}{r} \quad\Rightarrow\quad g = \frac{GM}{r^{2}}\$

6. Thus, for a circular orbit the relationship g = GM/r can be used when solving for orbital speed or period.

Practical Applications

• Calculating the acceleration due to gravity at the surface of a planet or moon.
• Determining orbital velocities of satellites.
• Estimating the weight of an object at different altitudes.

Worked Example

Problem: Find the gravitational field strength at a distance of 4.0 × 10⁶ m from the centre of a planet with mass 5.0 × 10²⁴ kg.

Solution:

1. Write the formula: \$g = \frac{GM}{r^{2}}\$
2. Substitute the known values:

   \$g = \frac{(6.674 \times 10^{-11}\,\text{N·m}^{2}\text{kg}^{-2})(5.0 \times 10^{24}\,\text{kg})}{(4.0 \times 10^{6}\,\text{m})^{2}}\$

3. Calculate the denominator: \$(4.0 \times 10^{6})^{2} = 1.6 \times 10^{13}\,\text{m}^{2}\$
4. Calculate the numerator: \$6.674 \times 10^{-11} \times 5.0 \times 10^{24} = 3.337 \times 10^{14}\$
5. Divide:

   \$g = \frac{3.337 \times 10^{14}}{1.6 \times 10^{13}} \approx 20.9\ \text{m s}^{-2}\$

6. Result: The gravitational field strength at that distance is approximately 20.9 m s⁻².

Common Mistakes to Avoid

• Confusing the radius r (distance from the centre) with the height above the surface.
• Omitting the square on r when using \$g = GM/r^{2}\$.
• Using the mass of the test object m in the expression for g; g depends only on the source mass M and distance r.
• Mixing units – ensure G, M, and r are all in SI units.

Summary Table of Symbols

Practice Questions

1. Calculate the value of g at an altitude of 300 km above the Earth’s surface. (Take Earth’s radius as 6.37 × 10⁶ m.)
2. A satellite orbits a planet of mass 8.0 × 10²⁴ kg at a radius of 2.0 × 10⁷ m. Determine the orbital speed using the relationship v = √(GM/r).
3. Explain why the gravitational field strength inside a uniform solid sphere varies linearly with distance from the centre.

Further Reading

• Newton’s law of universal gravitation – derivation and limitations.
• Kepler’s laws and their connection to \$g = GM/r\$.
• Gravitational potential energy and escape velocity.