why

Gravitational Field

Definition

Definition: Gravitational Field

Force per unit mass

Gravitational Field as a Field of Force

• A Gravitational Field is an example of a Field of Force.
• It is a region of space where a mass experiences a force.

Nature of the Force

• Gravitational forces are always attractive.
• The force always acts towards the centre of mass of the mass creating the field.

Representing Gravitational Fields

Gravitational Field Lines

• A gravitational field can be represented by using Gravitational Field Lines.
• A gravitational field line is an arrow showing the direction of gravitational force acting on a test mass.
• The density (number of field lines per unit area) of field lines represents the strength of the gravitational field; The closer the field lines the stronger the field.

Types of Gravitational Fields

Radial Field

• Field lines point towards the centre of a point mass or uniform sphere.

Gravitational field around the Earth

Spacing of Field Lines

As the distance from the centre of the point mass increases, the separation of the field lines increases, and line density decreases, indicating that the strength of the gravitational field decreases.

Uniform Field

• Field lines are parallel to each other and equally spaced

Gravitational field near the Earth's surface

Why is gravitational field considered to be uniform near earth’s surface

Important

• At surface, gravitational field lines are radial.
• Height/change in height above surface is very much smaller than Earth’s radius ($h\ll R$).
• field lines are approximately parallel.
• Field strength is represented by the density of field lines.
• Since field lines are parallel and equally spaced, density and therefore field strength is constant.

Gravitational force between point masses

Point Mass Approximation

The Point Mass Approximation

For a point outside a uniform sphere (like a planet), the mass of the sphere can be considered to be completely concentrated as a point mass at its centre. Condition: This is only valid if the sphere has uniform density (or spherical symmetry) and the distance $r$ is greater than the radius of the sphere.

Newton’s Law of Gravitation

Newton's Law of Gravitation

• The gravitational force of attraction between two point masses is directly proportional to the product of their masses
• and inversely proportional to the square of their separation.

Representation of M, m, r, F

Derivation

$$\begin{array}{rl}F& \propto \frac{Mm}{r^2}\\ & \\ F& =\frac{GMm}{r^2}\end{array}$$

Where $G$ is the Gravitational Constant, $G=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-2}$

Circular Orbits

Satellites

Satellites

A satellite is a mass that orbits (moves in circular motion) around a larger mass.

The Earth is a satellite orbiting the Sun and the Moon is a satellite orbiting the Earth.

How does a satellite stay in orbit?

• The gravitational force between the satellite and the mass which it orbits provides the centripetal force needed to keep the satellite moving in circular motion around the mass.

Gravitational Force as Centripetal Force

A satellite of mass m orbiting a mass of mass M

Orbital Velocity

Derivation of Orbital Velocity ( $v$)

• Explicitly state that:

$$\text{Gravitational Force}(F_g)\text{provides}\text{Centripetal Force}(F_c)$$

• Equate Gravitational Force and Centripetal Force:

$$\begin{array}{rl}{F}_g& =F_c\\ \frac{GMm}{r^2}& =\frac{m{v}^2}{r}\end{array}$$

• Simplify and make $v$ subject:

$$\begin{array}{rl}\frac{GMm}{r^2}& =\frac{m{v}^2}{r}\left(\times \frac{r}{m}\right)\\ \frac{GM}{r}& =v^2\\ & \\ v& =\sqrt{\frac{GM}{r}}\end{array}$$

Orbital Period

Derivation of Orbital Period ( $T$)

• Explicitly state that:

$$\text{Gravitational Force}(F_g)\text{provides}\text{Centripetal Force}(F_c)$$

• Equate Gravitational Force and Centripetal Force:

$$\begin{array}{rl}{F}_g& =F_c\\ \frac{GMm}{r^2}& =mr{\omega }^2\end{array}$$

• Simplify:

$$\begin{array}{rl}\frac{GMm}{r^2}& =mr{\omega }^2\left(\times \frac{1}{mr}\right)\\ \frac{GM}{r^3}& ={\omega }^2\end{array}$$

• Substitute $\omega =\frac{2\pi }{T}$:

$$\begin{array}{rl}\frac{GM}{r^3}& ={\left(\frac{2\pi }{T}\right)}^2\end{array}$$

• Simplify and make $T$ subject:

$$\begin{array}{rl}\frac{GM}{r^3}& =\frac{4{\pi }^2}{T^2}\\ T^2& =\frac{4{\pi }^2{r}^3}{GM}\\ T& =\sqrt{\frac{4{\pi }^2{r}^3}{GM}}\\ & \\ T& =2\pi \sqrt{\frac{r^3}{GM}}\end{array}$$

Kepler's Third Law of Planetary Motion

$T^2\propto r^3$

Geostationary Orbit

Geostationary Orbit

A geostationary orbit is one in which the orbiting satellite is always above the same point on Earth.

A telecommunications satellite orbiting the Earth

Conditions for Geostationary Orbit

• Orbital period of exactly 24 hours.
• Orbits from West for East.
• Orbits directly above the Equator.

Binary Star System

Two stars, A and B orbiting a common centre of gravity.

Newton's Third Law Pair

• According to Newton’s Third Law, Star A exerts a force on Star B, and Star B exerts an equal and opposite force on Star A.
• Star A experiences a force of magnitude $F$ towards the centre of Star B.
• Star B experiences a force of magnitude $F$ towards the centre of Star A.

Centripetal Force

• The gravitational force between the two stars provides the centripetal force for each star.
• Since the centripetal force of each star is equal to the gravitational force between them, their centripetal forces are equal in magnitude.

Barycentre

• The two stars orbit around their common centre of gravity, known as the barycentre

Conditions for a Binary Star System

• Same orbital period.
• Same angular velocity.
• Orbit in the same direction (clockwise or anticlockwise).
• Always opposite to each other across the barycentre.

Mathematical Descriptions of a Binary Star System

Since centripetal force of each star is equal:

$$\begin{array}{rl}{F}_{c_A}& =F_{c_B}\\ m_1{r}_1{\omega }^2& =m_2{r}_2{\omega }^2\left(\times \frac{1}{{\omega }^2}\right)\\ m_1{r}_1& =m_2{r}_2\\ & \\ \frac{m_1}{m_2}& =\frac{r_2}{r_1}\end{array}$$

Taking mass and orbital radius of the other star to be constant, the mass of star is inversely proportional to orbital radius:

$$\begin{array}{rl}{m}_1& =\frac{m_2{r}_2}{r_1}\\ m_1& \propto \frac{1}{r_1}\end{array}$$

Hence, the barycentre will be closer to the more massive star

Danger

• When using centripetal force equations, the radius is the distance from the star to the barycentre, not the distance between the stars.

Since gravitational force provides centripetal force

$$\begin{array}{rl}\text{Gravitational Force}(F_g)& \text{provides Centripetal Force}(F_c)\\ & \\ F_g& =F_c\end{array}$$ $$\begin{array}{rl}\frac{G{m}_1{m}_2}{(r_1+r_2{)}^2}& =m_1{r}_1{\omega }^2\\ \frac{G{m}_2}{(r_1+r_2{)}^2}& =r_1\times {\left(\frac{2\pi }{T}\right)}^2\\ \frac{G{m}_2}{(r_1+r_2{)}^2}& =r_1\times \frac{4{\pi }^2}{T^2}\\ T& =2\pi (r_1+r_2)\sqrt{\frac{r_1}{G{m}_2}}\end{array}$$

Let $R=r_1+r_2$

$$r_2=R-r_1$$ $$m_1{r}_1=m_2(R-r_1)$$ $$m_1{r}_1=m_{2}R-m_2{r}_1$$ $$r_1(m_1+m_2)=R{m}_2$$ $$r_1=\frac{R{m}_2}{m_1+m_2}$$

Gravitational Potential

Definition: Gravitational Potential at a Point

Work done per unit mass in bringing a small test mass from infinity to the point

Gravitational Potential at a Point ( $\varphi$)

$\varphi =-\frac{GM}{r}$

SI Unit: $\mathrm{Jk}{\mathrm{g}}^{-1}$

Important

• Gravitational potential is always negative
• Gravitational potential is zero at infinity ($-\frac{GM}{\infty }=0$)

Derivation of Gravitational Potential

Recall that work done is defined as: $W=F\times d$

Gravitational potential ($\varphi$) is defined as the work done per unit mass:

$$\begin{array}{rl}\varphi & =\frac{F\times d}{m}\\ & =\frac{GMm}{r^2}\times \frac{r}{m}\\ & =\frac{GM}{r}\end{array}$$

Since $\varphi =0$ at infinity, the potential is defined as:

$\varphi =-\frac{GM}{r}$

Why is Gravitational Potential Always negative?

• Gravitational Potential is defined as zero at infinity.
• Gravitational force is always attractive.
• Work is done on a mass to move it to infinity.
• Therefore potential must be negative at all points closer than infinity.

Work done to bring mass to infinity

Gravitational Potential Energy

Definition

$\text{Gravitational Potential Energy}=\text{Gravitational Potential}\times \text{mass}$

Derivation of Gravitational Potential Energy ( $E_p$)

$$\begin{array}{rl}{E}_p& =\varphi \times m\\ E_p& =-\frac{GM}{r}\times m\\ E_p& =-\frac{GMm}{r}\end{array}$$

Derivation of Change in Gravitational Potential Energy ( $\Delta E_p$)

$$\begin{array}{rl}\Delta E_p& =\Delta \varphi \times m\\ \Delta E_p& =\left(-\frac{GM}{r_{final}}-\left(-\frac{GM}{r_{initial}}\right)\right)\times m\\ \Delta E_P& =\left(\frac{GM}{r_{initial}}-\frac{GM}{r_{final}}\right)\times m\\ \Delta E_p& =GM\left(\frac{1}{r_{initial}}-\frac{1}{r_{final}}\right)\times m\end{array}$$