Newton's Law of Gravitation

The diagram above shows two masses with a separation of $r$.

We can say that $F_1 = F_2 = F_G$. We can write the gravitational force as:

$F_G = dfrac{Gm_1m_2}{r^2}$

• $F_G$ is the gravitational force between two masses in Newtons $left(text{N}right)$.
• $G$ is Newton’s gravitational constant, which is $6.67 times 10^{-11} : text{Nm}^2 text{kg}^{-2}$.
• $r$ is the separation of the two masses in metres $left(text{m}right)$.
• $m_1$ and $m_2$ are the masses of each object in kilograms $left(text{kg}right)$.

It is important that the separation is measured from the centre of each point mass rather than the surface.

As the force due to gravity between the two objects is inversely proportional to the separation squared, we can conclude that the force between the two objects follows the inverse square law. Therefore doubling the distance between the two objects causes the force to decrease by four times etc.

Example: A satellite is orbiting the Earth at a distance of $2500$ metres above the surface of the Earth. Given the below information, calculate the force due to gravity on the satellite.

Mass of Satellite $= 5000 : text{kg}$

Mass of the Earth $= 5.97 times 10^{24} : text{kg}$

Radius of Earth $= 6.37 times 10^6 : text{m}$

Gravitational constant $left(Gright) = 6.67 times 10^{-11} text{Nm}^2  text{kg}^{-2}$

[4 marks]

Gravitational potential is always negative as work is done against gravity as the body moves from the surface of the Earth to infinity. The gravitational potential will always be maximum at the surface and decrease to zero at infinity.

Gravitational potential can be calculated using the equation:

$V = dfrac{-GM}{r}$

• $V$ is the gravitational potential in joules per kilogram $left(text{Jkg}^{-1}right)$.
• $G$ is Newton’s gravitational constant, which is $6.67 times 10^{-11} : text{Nm}^2 text{kg}^{-2}$.
• $M$ is the mass of the body producing the force in kilograms $left(text{kg}right)$.
• $r$ is the separation between the body causing the force and the point in metres $left(text{m}right)$.

Example: A planet of mass $1.2 times 10^{23} : text{kg}$ has a diameter of $5000 : text{km}$. A large piece of rock is accelerated towards the planet and has a mass of $600 : text{kg}$. What would the gravitational potential of the rock be at a distance of $500 : text{km}$ from the surface of the planet? 

[3 marks]

$r = dfrac{5000}{2} + 500 = 3000 : text{km} = 3 times 10^6 : text{m}$

$V = dfrac{-GM}{r}$

$V = dfrac{left(-6.67 times 10^{-11} times 1.2 times 10^{23}right)}{3 times 10^6}$

$V = boldsymbol{-2.7 times 10^6} : textbf{Jkg}^{-1}$

Gravitational Potential Difference

Gravitational potential difference $left(Delta Vright)$ can be used to describe the difference in gravitational potential $left(Vright)$ between two points within the gravitational field.

$Delta V = V_f – V_i$

where $V_f$ is the final gravitational potential and $V_i$ is the initial gravitational potential, both in joules per kilogram $left(text{Jkg}^{-1}right)$.