Gravitational fields (3.7.2)
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An object with mass generates a gravitational field around it. Objects with mass are attracted to each other: an object in a gravitational field is attracted to the source of that field.
The strength of a gravitational field depends on the mass of the object and the distance from the object.
The gravitational field strength decreases as the distance from the mass increases. It follows an inverse-square relationship:
This relationship holds true for all gravitational fields, regardless of the mass generating it. For example, doubling the distance from a mass decreases the gravitational field strength by a factor of four.
Gravitational field strength, (at a given distance) is directly proportional to the mass of the object.
Larger masses produce stronger gravitational fields at the same distance. For example, a person standing on the Earth’s surface experiences a gravitational force of towards the centre of the Earth.
On the other hand, the gravitational force between two electrons is a factor of less than the Coulomb force of repulsion between them, so can be ignored.
Newton’s shell theorem states that a spherical shell of mass exerts the same gravitational pull on external objects as if all its mass were concentrated at its centre.
Calculations related to gravitational fields of spherical objects can be simplified by treating their mass as concentrated at a single point at their centre – this is the point mass approximation.
This approximation holds for spherically symmetric objects where the mass distribution is uniform.
The point mass approximation is commonly used for planets, stars, and other celestial bodies to calculate gravitational effects on nearby objects.
For example, the Earth can be approximated to a point mass of at the centre.
Without this approximation, finding the gravitational force acting on an object due to the Earth would require summing up the effects from each point on the Earth, which would be extremely difficult.
Remember to use the point mass approximation when calculating gravitational field strength or gravitational force between masses.
The point mass approximation applies only at distances greater than the radius of the spherical mass.
Inside the spherical mass, the gravitational field strength is influenced by the mass distribution.
Inside a spherical mass, the gravitational field strength at a point depends on the amount of mass inside within the radius equal to the distance at that point.
As you move away from the centre, more mass is enclosed within that radius, so the gravitational field strength actually increases.
This increase turns out to be directly proportional to the distance from the centre, which explains why the beginning of the graph above is a straight-line.
At distances greater than the radius of the mass, the inverse-square relationship is observed.
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.
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, .
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.
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.
The point mass approximation can be used for spherically symmetrical objects, where the mass distribution is uniform.
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.
Use the point mass approximation when calculating gravitational field strength or gravitational force between masses.
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.
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 .
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.
In a gravitational field, the field lines:
- represent the direction and strength of a gravitational field,
- indicate the force on a mass at each point in a gravitational field,
- always point inward towards the mass, as gravity is always an attractive force, and
- never cross each other.
If the gravitational field lines are closer together, the field is stronger at that point. When the gravitational field lines are spaced further apart, the field is weaker. Gravitational field-line diagrams show that the field strength decreases with distance from the centre of mass.
A spherical mass produces a radial, symmetrical gravitational field, equivalent to a point mass at its centre. Non-spherical masses only approximate this pattern at large distances; close to the object, the field lines follow the actual shape of the mass distribution.
In a uniform gravitational field, the field lines are parallel and equidistant, indicating a constant gravitational field strength over the region.
The gravitational field close to the surface of a planet is approximately uniform. The gravitational field strength remains approximately constant over relatively small distances from the surface.
Although the planet’s gravitational field is radial, the distance between the test mass (the object experiencing gravity) and the centre of the planet does not change much relatively when you are near the surface of a planet.
Gravitational field lines always point inward towards the mass, and never away. This is because gravity is an attractive force. On the other hand, electric field lines can point inward or outward.
Gravitational field strength, , is the force per unit mass at a point in a gravitational field. It is equal to the force exerted on a mass at a point in a gravitational field.
The formula for gravitational field strength is:
Where:
- is the gravitational force acting on an object in the gravitational field in newtons (), and
- is the mass of the object in kilograms ().
The unit for gravitational field strength is newtons per kilogram . This is equivalent to metres per second squared which is the SI unit for acceleration.
The equivalence in units of gravitational field strength and acceleration arises from the following two equations:
While both are expressions of Newton’s second law, they serve distinct purposes:
- : applies to any object experiencing a force, regardless of the cause.
- : a specific application describing an object under the influence of a gravitational field.
The higher the mass and the smaller the radius of a celestial body, the greater the gravitational field strength at its surface.
Different celestial bodies have different values of , which refers to the gravitational field strength at the body’s surface. The gravitational field strength on a planet determines the force acting on an object or person due to gravity. The stronger the gravitational field strength, the heavier the object will feel, and the more force is needed to lift it.
Question walkthrough
Astronaut's Weight on Mars vs Earth
Converts an astronaut's weight on Earth into mass, then uses Mars's gravitational field strength to find their weight on Mars.
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.
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.
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.
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.
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%.
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.
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.
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.
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.
The gravitational potential at a point is the work done per unit mass to bring a test mass from an infinite distance to a chosen point inside a body’s gravitational field.
At an infinite distance away from an object, the gravitational potential is defined as zero:
A mass at this point feels no force due to the object’s gravitational field.
Gravitational potential is always negative because:
- Gravitational forces are attractive. Energy is required to move a mass away from another mass. Therefore, the work done to move a mass from infinity towards another mass is negative.
- At any point within a gravitational field, the potential is lower than at infinity (where ).
Gravitational potential can be thought of as a ‘gravitational well’. Moving a mass to infinity is like climbing out of the well – energy is needed to reach the ‘zero’ level. A larger mass has a stronger gravitational field, corresponding to a larger ‘gravitational well’.
Gravitational potential energy is the energy a mass has due to its position in a gravitational field. It can be found from by multiplying by the object’s mass :
Gravitational potential at a point is the gravitational potential energy per unit mass.
Gravitational field strength is given by:
Where:
- is Newton’s gravitational constant ,
- is the mass of the body producing the gravitational field (kg),
- is the distance (m) from the centre of the mass to the point in the field, and
- is measured in
Comparing this to the equation for gravitational potential shows that
Where:
- is the energy per unit mass at a point in a gravitational field.
- is the force per unit mass acting on an object in a gravitational field.
It is important to note that and are directly proportional to each other, but they are distinctly different quantities. Moreover, both quantities are generally measured from the body’s centre of mass. However, for practical applications, we often express these quantities at different distances from the centre, including the surface.
The zero point for gravitational potential is at a distance of infinity. Hence, all distances closer than infinity have negative gravitational potential – this reflects that work is required to escape a gravitational field.
The gravitational potential at a distance from a mass is given by:
Where:
- is Newton’s gravitational constant ,
- is the mass of the body producing the gravitational field ,
- is the distance in from the centre of the mass to the point in the field.
is negative because gravitational forces are attractive, and has units of
The closer a point is to the mass the more negative the gravitational potential. As increases, becomes less negative, approaching zero at infinity.
Work is required to move a unit mass away from a planet or mass This work increases as the object moves further away.
Two points at different distances from a mass have different gravitational potentials because increases (becomes less negative) with distance.
The gravitational potential difference between two points is equal to:
Where:
- is the final gravitational potential,
- is the initial gravitational potential,
- is the change in gravitational potential.
All quantities are in
The change in potential between two distances (initial) and (final) from a mass is:
Where:
- is the gravitational constant
- is the mass of the object creating the field ()
- is the initial distance from the mass ()
- is the final distance from the mass ()
The change in potential is measured in
Question walkthrough
Work Done Moving a Satellite in a Gravitational Field
Calculate the work done moving a satellite between two distances from a planet's centre using the change in gravitational potential.
Newton’s law of gravitation states that the force between two masses and is given by:
Where:
- is Newton’s gravitational constant
- and are the masses of the larger body and small body, respectively (),
- is the distance between the two masses ().
This equation shows gravitational force is inversely proportional to This means a force–distance graph will be a curve (not a straight line). As increases, decreases rapidly and approaches zero at large
Work done is the product of force and distance
In a force–distance graph, the area under the curve represents the work done (or energy transferred) to move an object between two points.
When a satellite (mass is moved from one point (A) to another (B) in a gravitational field:
- Gravity is attractive – work is required to move the satellite away from the source of the field.
- The work done increases the satellite’s gravitational potential energy.
- The area under the curve between points A and B represents the satellite’s change in gravitational potential energy.
Question walkthrough
Estimating Work Done Using a Constant-Force Approximation
Estimate the work done moving a spacecraft a small distance in a gravitational field by treating the gravitational force as approximately constant.
The gravitational potential energy of an object in a gravitational field is given by:
Where:
- is the object’s mass (kg),
- is the gravitational potential at that point
The equation for is:
Substituting this into the equation for gravitational potential energy gives:
Where:
- is the universal gravitational constant ,
- is the mass creating the gravitational field (),
- is the distance from the centre of mass to the object of mass (), and
- is measured in Joules ().
Gravitational potential energy is negative because the gravitational force is attractive, and energy is required to move a mass away from the source of the field. At infinity, gravitational potential energy is zero.
Work is required to move a mass against a gravitational field.
The work done or energy transferred ) to move an object between two points in the gravitational field, with a change in gravitational potential ) is given by:
Where:
- is the change in gravitational potential between two points in the field with gravitational potentials and
- is the object’s mass .
This work is equal to the change in gravitational potential energy of the mass. If there is no change in potential there is no change in This means no work is done moving an object between points with the same potential.
When a mass is moved from one point to another point in the gravitational field of a mass , the change in gravitational potential energy is:
Which can be written as:
Where:
- is the mass creating the gravitational field ()
- is the mass moving in the field ()
- and are the initial and final distances from the centre of mass ()
Moving a mass further from the source of the gravitational field (increasing results in a positive change in potential energy, meaning energy is required to move the mass.
Conversely, moving a mass closer to the source (decreasing results in a negative change in potential energy, releasing energy in the process.
Kepler’s first law of planetary motion is related to the shape of orbits.
Orbiting objects (such as planets, moons, or artificial satellites) move in elliptical orbits with the object they are orbiting (such as a star or planet) at one of the two foci.
It is useful to know that all the planets in our solar system (including Earth) orbit in elliptical paths with low eccentricity, meaning that the orbits are approximately circular.
Kepler’s second law of planetary motion is related to the movement of planets.
A line joining an orbiting object and the object that it is orbiting sweeps out equal areas in equal times
This implies that orbiting objects move faster when closer to the sun (perihelion) and slower when farther from the Sun (aphelion).
Kepler’s second law results from the conservation of angular momentum, where the orbiting body’s speed adjusts to maintain a constant motion around its host.
Kepler’s third law of planetary motion relates orbital time to radius.
The square of the orbital time period is directly proportional to the cube of the orbital radius:
Here is the semi-major axis of the elliptical orbit, which is the greatest distance between the planet and the centre of the elliptical orbit.
The third law mathematically demonstrates how the time for one orbit increases with distance from the body being orbited.
Kepler’s laws provide an excellent approximation of planetary motion using Newtonian physics. However, they do not fully account for the effects of general relativity.
An example of this is Mercury’s orbit of the Sun. It deviates slightly from Keplerian predictions due to the curvature of spacetime caused by the Sun’s immense gravitational field. This discrepancy, known as the precession of Mercury’s perihelion, was accurately explained by Einstein’s theory of general relativity.
Question walkthrough
Minimum Orbital Velocity Using Kepler's Second Law
Identify the point on an asteroid's elliptical orbit where its velocity is at a minimum, and explain using Kepler's second law.
The centripetal force acting on a planet is the gravitational force of the sun on the planet.
The direction of the centripetal force is towards the sun and perpendicular to the direction of motion of the planet.
It is important to note that for simplified models typically found at A level, the orbital path of most planets can be approximated as circular.
The centripetal force required to keep a planet in orbit is provided by the gravitational force between the planet and the Sun, such that:
Where:
- is the mass of the planet in
- is the orbital velocity of the planet in
- is the angular velocity of the planet in
- is the orbital radius (distance between the centre of masses of the planet and sun) in
- is the mass of the Sun, in
- is the gravitational constant
Using the relation between gravitational force and centripetal force the orbital velocity, , of a planet in a circular orbit can be derived:
The planet’s mass cancels out on both sides of the equation, meaning that the orbital velocity and centripetal force only depend on the Sun’s mass and the planet’s distance. This means that all planets, regardless of mass, travel at the same speed at a given orbital radius.
It is important to understand the role that force and acceleration play in circular motion.
Question walkthrough
Orbital Radius from Gravitational-Centripetal Force
Equate gravitational and centripetal force to calculate a planet's orbital radius from its speed and its star's mass.
The orbital time period and radius of an orbiting body are given by:
Where:
- is the orbital time period (time taken to complete one orbit) in
- is the orbital radius (distance between the centre of masses of the orbiting body and the body at the centre of the orbit) in
- is the mass of the Sun in
- is the gravitational constant = .
It is important to know the following derivation for your exams on how the orbital period and speed are related to the radius of a circular orbit. It assumes the central body (e.g. the Sun) dominates the gravitational field and no other planets are impacting the orbiting body.
The net force acting on an orbiting body is equal to the gravitational force of the host body:
The acceleration of the planet is perpendicular to its velocity, so it is purely centripetal:
Combine equations (1) and (2):
Angular speed is given by , where is the time taken for one full rotation (orbital time period):
Kepler’s third law applies to any satellite or moon orbiting a planet, provided the dominant mass is much greater than the orbiting object:
It is useful to note that if is at least 100 times larger than , the error in using Kepler’s Third Law without modification is generally very small (less than 1%).
Examples of systems where Kepler’s law can be applied, other than our solar system:
- Artificial satellites orbiting Earth
- Moons orbiting a planet
- Exoplanets orbiting other stars
Question walkthrough
Comparing Orbital Speeds at Different Radii
Use Kepler's third law to find the orbital speed of a planet at twice the orbital radius of another, in terms of the first planet's speed.
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Kepler's Third Law and Mass of Jupiter
Verify Kepler's third law using orbital data for three of Jupiter's moons, then use it to calculate the mass of Jupiter.
A geostationary orbit is a circular orbit around Earth in which a satellite remains fixed above the same point on the equator. The satellite’s orbital period is exactly 24 hours, matching Earth’s rotational period.
This is a special case of a geosynchronous orbit, where the orbital period is 24 hours, but the satellite does not necessarily stay above the same point.
Conditions for a geostationary orbit:
- The orbital period equals 24 hours; the satellite must complete one orbit at the same time that Earth completes one full rotation.
- The satellite moves from west to east (in the same direction as the Earth’s spin).
- For an equatorial orbit, the satellite must orbit directly above the equator (zero inclination), which means that it remains in the same position above the Earth’s surface.
- The orbit must be perfectly circular, not elliptical.
The table below lists some of the applications of geostationary satellites.
The table below lists the advantages and disadvantages of the use of geostationary satellites.
Question walkthrough
Orbital Speed of a Geostationary Satellite
Calculate the orbital speed of a geostationary satellite from its orbital radius and the Earth's rotation period.
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Geostationary vs Low Earth Orbit Satellites
State a use of geostationary satellites and compare their advantages and disadvantages against low Earth orbit satellites for that use.
Escape velocity is the minimum speed an object must travel to escape a gravitational field completely without any further energy input.
Escape velocity depends only on the mass creating the gravitational field and the position of the escaping object in the field; it does not depend on escaping the object’s mass.
The equation for escape velocity is:
Where:
- is the universal gravitational constant (measured in
- is the mass () creating the gravitational field,
- () is the distance from the centre of mass to the escaping object.
Whether a tennis ball or a spaceship, all objects have the same escape velocity in the same gravitational field (at the same position).
Remember that escape velocity is the minimum speed an object must travel to completely escape a gravitational field without any further energy input.
Lighter gas molecules (e.g. hydrogen and helium) can escape planetary atmospheres if their speeds exceed escape velocity.
The Maxwell–Boltzmann distribution gives the average speed of gas molecules, where is the root mean square speed. The is the square root of the average of the square speeds of all molecules in the system.
Where:
- is the Boltzmann constant,
- is the temperature in Kelvin,
- is the mass of the gas molecule in ()
If the average molecular speed approaches the escape velocity, the planet will gradually lose that gas.
For example:
- Earth retains oxygen and nitrogen (heavier gases move more slowly).
- Small planets (e.g. Mars) and moons have lost most of their atmospheres.
- Due to their high escape velocities, Jupiter and Saturn retain light gases like hydrogen.
Question walkthrough
Deriving the Escape Velocity Formula
Derive the escape velocity formula from the kinetic and gravitational potential energy equations.
Question walkthrough
Comparing Escape Velocity and Gas Particle Speed
Compare an exoplanet's escape velocity with the Maxwell–Boltzmann RMS speed of helium atoms to determine whether the atmosphere is retained.