The law of gravitation
In a nutshell
Newton's law of gravitation gives the gravitational force acting between two separate masses separated by a distance. Newton's law of gravitation can also be used to find the gravitational field strength of a point mass at a distance away. Both equations are inverse square laws which means that the values of force and gravitational field strength decreases exponentially with distance.
Equations
Description | Equation |
Newton's law of gravitation | F=(−)Gr2Mm |
Gravitational field strength of a point source | g=(−)Gr2M |
Constants
name | symbol | value |
gravitational constant | | 6.67×10−11Nm2kg−2 |
Variable definitions
Quantity Name | Symbol | Derived Unit | SI BASE Units |
gravitational force | | | kgms−2 |
larger point mass | | | |
smaller point mass | | | |
distance between centres of masses | | | |
gravitational field strength | | | |
Newtons law of gravitation
Newtons law of gravitation is an equation which is used to calculate the force acting on a point mass within a gravitational field.
The equation is given as:
F=−Gr2Mm
G is the gravitational constant and is given in the exam data booklet. The equation has a minus sign in it because the gravitational force is always attractive and hence is negative.
Note: M denotes the mass of the object which is creating the gravitational field and m is the point mass within the field. This varies depending on context.
Example
When considering the force of the Earth on the Moon, M would be the mass of the Earth as it is creating the gravitational field and m is the mass of the Moon.
However if considering the gravitational force of the Sun on the Earth, M would be the mass of the Sun and m would be the mass of the Earth.
Example
Calculate the gravitational force on a Starlink satellite which has a mass of 1250kg and orbits at a distance of 530km above Earth's surface.
REarth=6.37×106mMEarth=5.97×1024kg
Firstly, write down the known values:
M=5.97×1024kgm=1250kgrorbit=530km=530000mREarth=6.37×106m
Next, write down the equations needed and rearrange if necessary:
F=−G(REarth+rorbit)2Mm
Note: The orbit of the satellite is 530km above the surface of the Earth, so the distance of the satellite to the centre of the Earth is REarth+rorbit
Then, substitute the values into the equation:
F=−6.67×10−11×(6.37×106+530000)25.97×1024×1250F=−10454...
Make sure to include units and round to the lowest number of significant figures of the values given by the question:
gravitational force, F=(−)10500N
The gravitational force of the Earth on the Starlink satellite is (−)10500N.
Note: Technically the value of the force is negative, but you won't get penalised in an exam if you state the force a positive magnitude of force.
The inverse square law
The gravitational force is inversely proportional to the distance squared of the point mass within a gravitational field.
This means that the gravitational force is strongest at the surface of a planet or star and decreases exponentially until it gets to 0N at an infinite distance away.
Example
The Starlink satellite has a gravitational force of roughly 10000N at a distance of 6.9×106m. If another satellite with the same mass were at twice the distance away, at 1.38×107m, the gravitational force would be 4 times less at roughly 2500N.
Mathematically this can be shown:
F∝r21
If distance r is doubled, then:
F∝221F∝41
The force is a quarter of the original value, or 4 times less.
Gravitational field strength
Gravitational field strength is defined as:
g=mF
Substituting Newton's law of gravitation for F:
g=−mr2GMmg=−r2GM
This equation can be used to find the gravitational field strength of a point mass at a distance of r metres away from the point mass.
Note: Similar to before, the mass in this equation is the object which is causing the gravitational field. (The most massive object in the scenario).
Graphically, how the gravitational field strength of a point mass varies with the distance away from the point mass looks like this:
It is another inverse square relationship.
Example
The following graph is for Jupiter.
The radius of Jupiter is 69900km. Calculate the mass of Jupiter.
Firstly, write down the known values:
rJupiter=69900km=69900000m
The graph can be used to obtain the gravitational field strength at the surface of Jupiter:
gJupiter=24.8Nkg−1
Next, write down the equations needed and rearrange if necessary:
g=r2GM→M=Ggr2
Then, substitute the values into the equation:
M=6.67×10−1124.8×699000002M=1.81669...×1027 kg
Make sure to include units and round to the lowest number of significant figures of the values given by the question:
mass of Jupiter, M=1.82×1027kg
The mass of Jupiter is 1.82×1027kg.