Module 5: Newtonian world and astrophysicsNewton's law of gravitation (5.4.2)

Newton's law of gravitation (5.4.2)

Universal gravitation, F = GMm/r², gravitational field strength, g = GM/r², radial and uniform fields in A-level Physics.
6 min

Newton’s law of gravitation states that the gravitational force between two point masses, separated by a distance, is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Newton’s law of gravitation can be expressed mathematically as:

Where:

  • is the gravitational force in newtons .
  • and are the masses of the two point bodies in .
  • is the separation distance in metres .
  • is the gravitational constant, equal to

The negative sign indicates that gravitational force is attractive – gravity pulls masses toward each other.

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The gravitational force acting on objects on the surface of the Earth and other planets is known as weight.

The weight of an object near a planet’s surface is given by:

Where:

  • is the object’s mass, and
  • is the gravitational field strength on the Earth’s surface, which is .

Different celestial bodies have different gravitational field strengths at their surface, depending on their mass and radius.

Since weight is equivalent to the gravitational force, it has the same unit, the newton, .

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The gravitational force between two masses decreases as the distance between them increases, following an inverse-square relationship:

The inverse-square relationship holds true regardless of the specific masses involved. For example, doubling the distance between any two masses decreases the gravitational force by a factor of four.

A graph showing the relationship between gravitational force, F (N), and distance from the centre of a spherical object, r (m). The curve decreases as the distance increases. The radius of the object is indicated as R (m).

The graph above shows the inverse-square relationship between the gravitational force, , and the distance from the centre of a spherical object, .

It is important to note that the curve is initially very steep and then becomes shallower. The gravitational force exerted by a mass (like a planet or star) diminishes rapidly as an object moves further away. However, this rate of decrease lessens significantly at greater distances.

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The point mass approximation can be used for spherically symmetrical objects, where the mass distribution is uniform.

An illustration showing the Earth on the left and the Moon on the right, with a dashed line connecting them labeled 'r'.

In Newton’s law of gravitation, the point mass approximation states that the separation distance between two masses is equal to the distance between their centres rather than the distance between their surfaces.

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Use the point mass approximation when calculating gravitational field strength or gravitational force between masses.

An illustration showing the Earth on the left and the Moon on the right, connected by a horizontal line labeled 'r'.
Do

Use the distance between the centres of the masses.

An illustration showing the Earth on the left and the Moon on the right, with a dashed line connecting them labeled 'r'.
Don't

Use the distance between the surfaces of the masses.

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When multiple objects are present, the total gravitational force acting on any single object is determined by summing the vectors of the individual gravitational forces exerted by all the other objects.

  • One-dimensional cases: If objects are aligned along a straight line, the magnitudes of the gravitational forces can be added by considering their directions.
  • Two-dimensional cases: For objects positioned at angles (e.g. triangular configurations), individual forces can be resolved into components using Pythagoras’ theorem or trigonometric functions (sine and cosine). Then, sum the components along each axis to find the net force.
A diagram illustrating three masses m1, m2, and m3. The force vectors are shown as F→12 in blue, F→13 in red, and F→1 in black. The distances are represented as r→12 and r→13.

The diagram above shows a triangular arrangement of masses. The total gravitational force on mass is equal to the vector addition of the gravitational forces and due to each of the masses and .

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Question walkthrough

Resultant Gravitational Force on a Collinear Mass

Calculate the resultant gravitational force on a mass from two other masses aligned along the same straight line.

Gravitational field strength, is the force per unit mass at a point in a gravitational field. It equals the force exerted on a mass at a point in a gravitational field.

Gravitational field strength is expressed mathematically as:

Where

  • is the gravitational force in newtons (N), and
  • is the object’s mass in the gravitational field in kilograms ().

The units for gravitational field strength are (or ), the same as for acceleration.

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Combining the formulas for gravitational field strength and Newton’s law of gravitation leads to an expression for gravitational field strength .

Substituting Newton’s law for gravitation:

into the equation for gravitational field strength:

gives:

Simplifying the expression gives the gravitational field strength, , at a distance from the centre of an object of mass as:

The negative sign indicates that the gravitational field strength at a distance is in the opposite direction to the displacement from the centre of mass , showing that the gravitational force is attractive, since is force per unit mass.

Note that the gravitational field strength does not depend on the object’s mass in the gravitational field.

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The magnitude of gravitational field strength , at a given and constant distance, is directly proportional to the mass of the object creating the gravitational field.

Moreover, the magnitude of the gravitational field strength decreases as the distance from the mass increases, following an inverse-square relationship.

Gravitational field strength at distance of 1 m, g (Nkg⁻¹) plotted against Mass, M (kg) in the upper graph, showing a linear increase. The lower graph shows Gravitational field strength, g (Nkg⁻¹) plotted against Distance from centre of spherical object, r (m), illustrating a decreasing curve with a dashed line indicating the Radius of object, R (m). © Medify
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A planet’s gravitational field is radial. However, for most everyday applications near the surface, the change in distance from the planet’s centre is negligible compared to the planet’s radius.

The gravitational field strength remains approximately constant over minor height differences. This means the field lines can be approximated as parallel and equidistant, making the gravitational field appear uniform near the surface.

A diagram showing a purple planet labeled 'Planet' with a diameter of '6368 km'. Arrows are pointing outward from the planet. There is a red 'A' marked on the planet and a corresponding 'A:' with vertical lines and arrows on the right side.

The gravitational field strength at the surface of Earth is .

On Earth, the difference in gravitational field strength at sea level and the peak of Mount Everest, only above sea level, is negligible. The field strength at the highest peak on Earth is , a percentage difference of only 0.4%.

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Compare the formula for gravitational field strength:

With Newton’s second law of motion:

This shows that gravitational field strength is equal to the acceleration of an object under gravity (and no other forces):

An object in free fall is accelerating under gravity and no other forces.

At the Earth’s surface, where is approximately constant, objects fall with the same acceleration regardless of their mass. However, this only applies in the absence of air resistance, which would cause an additional force.

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In 1971, astronaut David Scott demonstrated that in the absence of air resistance, objects fall at the same rate under gravity, regardless of their mass.

Scott dropped a hammer and a feather on the surface of the Moon, and they hit the ground simultaneously.

An astronaut in a space suit stands on a surface with a blue background. To the left of the astronaut is a red hammer, and to the right is a yellow object that resembles a feather.

On Earth, air resistance slows objects with larger surface areas, so a feather falls more slowly than a hammer.

On the other hand, the Moon has no atmosphere, so there is no air resistance. As a result, both the feather and hammer fall at the same rate.

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The acceleration due to gravity is the same for all objects in free fall, provided air resistance is negligible. In the presence of a fluid, acceleration can vary due to differences in surface area and shape and air (or fluid) resistance.

A comparison image showing two scenarios: on the left, labeled 'In air', a feather and a hammer are depicted, and on the right, labeled 'In a vacuum', the same feather and hammer are shown.
Do

Remember that in the absence of air resistance, all objects fall at the same rate under gravity, regardless of their mass.

For example, when a hammer and a feather are dropped on the moon from the same height and at the same time, they hit the surface simultaneously.

The image shows two panels. The left panel is labeled 'Air resistance present' and depicts a feather falling slowly. The right panel is labeled 'No resistance present' and shows a hammer falling quickly.
Don't

Assume heavier objects always fall faster.

A feather and a hammer will only fall at different rates when air resistance is present.

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