# Newton's Law of Gravitation

## Newton's Law of Gravitation Revision

**Newton’s Law of Gravitation and Gravitational Potential**

Sir Isaac Newton made a massive impact in our understanding of physics. One of his famous contributions to physics was the idea of **gravitation**. In this section we look at **Newton’s law of gravitation** and **gravitational potential**.

**Newton’s Law of Gravitation**

Isaac Newton aimed to calculate the force of **gravity** acting between two objects. For his calculations, he modelled the objects as point masses. His law of **gravitation **can be applied to calculate the **gravitational force** between two point masses (any spherical object like planets or stars).

**Newton’s law of gravitation** states that any object with a mass will attract another object with a mass proportionally to the product of their **mass** and** inversely proportionally** to their separation squared. Therefore:

F \propto m_1 m_2

F \propto \dfrac{1}{r^2}

This results in an equation we will discuss in the next section.

**Calculating Gravitational Force Between Two Objects**

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(G\right) = 6.67 \times 10^{-11} \text{Nm}^2 \text{kg}^{-2}

**[4 marks]**

To understand this question further, draw a diagram of what we know from the question:

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

F_G = \dfrac{\left(6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 5000 \right)}{\left(2500 + 6.37 \times 10^6\right)^2}

F_G =\boldsymbol{ 4.9 \times 10^4} \: \textbf{N} \: \text{or} \: \boldsymbol{49}\textbf{kN}

**Gravitational Field Strength in a Radial Field**

Previously we learnt how to calculate the **gravitational field** strength on an object. However, we also discussed that the **gravitational field strength** varies with distance away from the object causing the **gravitational force**.

To calculate the the **gravitational field strength** \left(g\right) at any point within a** radial field**, the following equation can be applied:

g = \dfrac{GM}{r^2}

- g is the
**gravitational field strength**in Newtons per kilogram or metres per second squared \left(\text{Nkg}^{-1} \: \text{or} \: \text{ms}^{-2}\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 field in kilograms \left(\text{kg}\right). - r is the
**separation between the body**causing the force and the point you are trying to calculate in metres \left(\text{m}\right).

**Example: **What would the mass of the moon need to be to produce a gravitational field strength at its surface the same as the Earth’s?

Radius of moon = 1.74 \times 10^6 \text{m}

Gravitational constant \left(G\right) = 6.67 \times 10^{-11} \: \text{Nm}^2\text{kg}^{-2}

Gravitational field strength at the Earth’s surface = 9.81 \: \text{Nkg}^{-1}

**[3 marks]**

g = \dfrac{GM}{r^2}

This rearranges to give:

M = \dfrac{gr^2}{G}

g on earth is 9.81 \: \text{Nkg}^{-1}. So we use this in our equation:

M = \dfrac{\left(9.81 \times 1.74 \times 10^6\right)}{\left(6.67 \times 10^{-11}\right)}

M = 4.45 \times 10^{23} \: \text{kg}

Therefore the moon would have to have a mass of 4.45 \times 10^{23} \: \text{kg} to produce a gravitational field of 9.81 \: \text{Nkg}^{-1} at it’s surface.

**Gravitational Potential**

The **gravitational potential energy **can be described as the amount of energy a body has due to its position within a **gravitational field**.

The **gravitational potential** \left(V\right) is another term commonly used and can be defined as the **work done per unit mass** when moving a mass at rest from **infinity** to a given point. In this definition, **infinity **is a point outside the **gravitational** **field**.

**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}

The graph represents the relationship between **gravitational potential** \left(V\right) and distance from the centre of the planet \left(r\right).

It shows how **gravitational potential** is always negative and it approaches zero at infinity.

**Gravitational Potential Difference**

**Gravitational potential difference** \left(\Delta V\right) can be used to describe the difference in** gravitational potential** \left(V\right) 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).

## Newton's Law of Gravitation Example Questions

**Question 1:** Describe Newton’s law of gravitation.

**[2 marks]**

Newton’s law of gravitation states that **any object with a mass will attract another object with a mass, proportionally to the product of their mass and inversely proportionally to their separation squared**.

OR

\boldsymbol{F_G = \dfrac{Gm_1m_2}{r^2}}

**Question 2:** A body is orbiting the Earth at a distance of 5700 \: \text{m} above the surface of the Earth. Given the below information, calculate the force due to gravity on the satellite.

Mass of Satellite = 8000 \: \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(G\right) = 6.67 \times 10^{-11} \: \text{Nm}^2 \text{kg}^{-2}

**[4 marks]**

**Question 3:** Describe the relationship between radius and gravitational potential.

**[1 mark]**

As the **radius increases, the gravitational potential also increases as it approaches zero.**

**Question 4:** A planet of mass 3.9 \times 10^{23} \: \text{kg} has a diameter of 7000 \: \text{km}. A large piece of rock is accelerated towards the planet and has a mass of 1000 \: \text{kg}. What would the gravitational potential of the rock be at a distance of 200 \: \text{km} from the surface of the planet?

**[3 marks]**

\begin{aligned} r &= \dfrac{7000}{2} + 1000 = 4500 \: \text{km} = 4.5 \times 10^6 \: \text{m} \\ V &= \dfrac{-GM}{r} \\ V &= \dfrac{\left(-6.67 \times 10^{-11} \times 3.9 \times 10^{23}\right)}{4.5 \times 10^6} \\ V &= \boldsymbol{-5.8 \times 10^6} \: \text{Jkg}^{-1}\end{aligned}