3.1 Newton's Universal Law of Gravitation
Gravitational force is a universal force that acts between any two bodies in the universe. In 1687, Sir Isaac Newton formulated the Universal Law of Gravitation after observing that an apple falls vertically to the ground and that the Moon revolves around the Earth due to mutual attraction.
Learning Objectives
- Explain Newton's Universal Law of Gravitation: \(F = \dfrac{Gm_1m_2}{r^2}\).
- Solve problems involving Newton's Universal Law of Gravitation for bodies at rest, satellites, and astronomical bodies.
- Relate gravitational acceleration, \(g\), on the surface of the Earth with the universal gravitational constant, \(G\), using \(g = \dfrac{GM}{r^2}\) and analyze the variation of \(g\) with distance, \(r\).
- Justify the importance of knowing the value of gravitational acceleration in space exploration and continuity of life.
- Explain centripetal force in the motion of satellites and planets using \(F = \dfrac{mv^2}{r}\) and centripetal acceleration using \(a = \dfrac{v^2}{r}\).
- Determine the mass of the Earth and the Sun using Newton's Universal Law of Gravitation and centripetal force (\(M = \dfrac{4\pi^2r^3}{GT^2}\)).
Graph of Variation of Gravitational Acceleration, \(g\), with Distance, \(r\), from Centre of Earth
See how \(g\) increases linearly inside the Earth and decreases with the inverse square of distance outside the Earth.
Inside the Earth: \(r < R\)
\(g\) is directly proportional to \(r\). At the centre of the Earth, \(r = 0\), so \(g = 0\).
Outside the Earth: \(r \ge R\)
\(g\) is inversely proportional to \(r^2\). If \(g(R)\) is the surface value, then \(g(2R) = \frac{1}{4}g(R)\) and \(g(3R) = \frac{1}{9}g(R)\).
Short Explanation
Newton's Universal Law of Gravitation
The gravitational force between two bodies is directly proportional to the product of the masses of both bodies and inversely proportional to the square of the distance between the centres of the two bodies.
Gravitational Force Formula
\(F = \dfrac{G m_1 m_2}{r^2}\)
\(F\) = gravitational force (N), \(m_1\) and \(m_2\) = masses (kg), \(r\) = distance between centres (m), \(G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
Gravitational Acceleration
On the Earth's surface, \(g = \dfrac{GM}{R^2} = 9.81\ \mathrm{m\,s^{-2}}\). At height \(h\), \(g = \dfrac{GM}{(R+h)^2}\).
Centripetal Force
For a satellite or planet in circular orbit, gravitational force provides the centripetal force: \(F = \dfrac{mv^2}{r}\), with centripetal acceleration \(a = \dfrac{v^2}{r}\).
Determining the Mass of the Earth or the Sun
\(M = \dfrac{4\pi^2 r^3}{G T^2}\)Equate gravitational force and centripetal force, then substitute \(v = \dfrac{2\pi r}{T}\).
Comparing Planetary Gravitational Acceleration
Surface gravitational acceleration differs because each body has a different mass and radius. Approximate values are the Moon \(1.62\ \mathrm{m\,s^{-2}}\), Mars \(3.71\ \mathrm{m\,s^{-2}}\), Earth \(9.81\ \mathrm{m\,s^{-2}}\), and the Sun \(274\ \mathrm{m\,s^{-2}}\). An astronaut's mass remains the same, but the astronaut's weight, \(W=mg\), changes.
Why Knowing \(g\) Matters
The value of \(g\) is needed to plan launches, landings, orbits, fuel, and mission payloads. Gravity also helps retain an atmosphere and influences the movement of fluids and living organisms. Long exposure to low gravity can reduce bone and muscle density, redistribute body fluids, and affect lung development and function, blood circulation, and blood pressure. Knowing \(g\) is therefore important for life-support systems and for human development and health during space exploration.
| Quantity | Symbol | SI Unit |
|---|---|---|
| Gravitational force | \(F\) | N |
| Mass of body | \(m\) | kg |
| Distance between centres | \(r\) | m |
| Gravitational constant | \(G\) | \(\mathrm{N\,m^2\,kg^{-2}}\) |
| Gravitational acceleration | \(g\) | \(\mathrm{m\,s^{-2}}\) |
| Linear speed | \(v\) | \(\mathrm{m\,s^{-1}}\) |
| Orbital period | \(T\) | s |
Try to Answer First
Answer in your mind, then press “Check Answer”.
Two identical spheres of mass \(5.0\ \mathrm{kg}\) each are separated by a distance of \(0.5\ \mathrm{m}\) between their centres. Calculate the gravitational force between them. (\(G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\))
Check Answer
What is the formula for the centripetal acceleration, \(a\), of a satellite orbiting Earth at linear speed \(v\) and orbital radius \(r\)?
Check Answer
What two physical measurements of an orbiting satellite are needed to determine the mass of the Earth?
Check Answer
A satellite of mass \(1.00\times10^3\ \mathrm{kg}\) is \(4.00\times10^5\ \mathrm{m}\) above Earth. Calculate the gravitational force on it if \(M=5.97\times10^{24}\ \mathrm{kg}\), \(R=6.37\times10^6\ \mathrm{m}\), and \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
Check Answer
The Moon orbits Earth with a mean radius of \(3.84\times10^8\ \mathrm{m}\) and a period of \(27.3\) days. Determine the mass of Earth.
Check Answer
Common Mistakes
- Using height above surface (\(h\)) instead of distance from centre (\(r = R + h\)) in gravitational force and gravitational acceleration calculations.
- Forgetting to square the distance (\(r^2\)) in \(F = \dfrac{Gm_1m_2}{r^2}\) and \(g = \dfrac{GM}{r^2}\).
- Assuming that because the Earth has a much larger mass, it exerts a larger gravitational force on a falling durian or satellite than the object exerts on the Earth. In fact, both bodies experience an attractive force of the exact same magnitude.
- Forgetting to convert orbital heights from kilometres (\(\mathrm{km}\)) to metres (\(\mathrm{m}\)) before substituting into the formulas.
Concept Misunderstandings
There is no gravity acting on satellites orbiting the Earth, which is why astronauts float.
Satellites and astronauts are within Earth's gravitational field. They experience free fall towards the Earth, and this gravitational force provides the centripetal force required to keep them in orbit.
Gravitational force only acts between planets and stars, not between ordinary people or small objects.
Gravitational force is universal and exists between any two masses. However, because \(G\) is extremely small, the force between everyday small masses is negligible and cannot be felt.
Gravitational acceleration increases continuously as you go deeper towards the Earth's centre.
Below the Earth's surface (\(r < R\)), gravitational acceleration decreases linearly with distance towards the centre and becomes zero at the centre of the Earth.
Summary
- Newton's Universal Law of Gravitation: \(F = \dfrac{Gm_1m_2}{r^2}\) acts between any two bodies; force is equal in magnitude and opposite in direction on both bodies.
- Gravitational constant: \(G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
- Gravitational acceleration: \(g = \dfrac{GM}{r^2}\) on the surface (\(r = R\), \(g \approx 9.81\ \mathrm{m\,s^{-2}}\)) and \(g = \dfrac{GM}{(R+h)^2}\) at height \(h\).
- Variation of \(g\): directly proportional to \(r\) inside the Earth (\(r < R\)), maximum at surface (\(r = R\)), and inversely proportional to \(r^2\) outside (\(r \ge R\)).
- Centripetal force and acceleration: \(F = \dfrac{mv^2}{r}\) and \(a = \dfrac{v^2}{r}\), directed towards the centre of the circular path.
- Orbital motion: gravitational force provides centripetal force (\(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\)).
- Mass of Earth/Sun: determined using orbital radius \(r\) and period \(T\) via \(M = \dfrac{4\pi^2r^3}{GT^2}\).
Short Activity
Objective: Identify gravitational force, gravitational acceleration, centripetal force and mass determination.
A. Objective Quiz
1 What happens to the gravitational force between two bodies if the distance between their centres is doubled while their masses remain unchanged?
2 What provides the centripetal force that keeps a man-made satellite in orbit around the Earth?
B. Fill in the Blanks
3 For a position above the surface of the Earth (\(r \ge R\)), the value of gravitational acceleration is inversely proportional to the of the distance from the centre of the Earth.
4 The formula for centripetal acceleration of a body moving at linear speed \(v\) in a circle of radius \(r\) is a = .
C. Matching / Drag and Drop
Drag each quantity to the correct formula or value. If using a phone, tap the answer first, then tap the matching box.
