Gravitation

Statement: Everybody in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Mathematical Derivation:
Consider two bodies of masses $m_1$ and $m_2$ separated by a distance $d$.
$F \propto m_1 m_2$
$F \propto \frac{1}{d^2}$
Combining both:
$F \propto \frac{m_1 m_2}{d^2}$
$F = G \frac{m_1 m_2}{d^2}$
Here, $G$ is the Universal Constant of Gravitation.

Ideally, the value of $G$ is the same everywhere in the universe. In SI units, its value is:
$G = 6.673 \times 10^{-11} Nm^2kg^{-2}$.
Because this value is very small, the gravitational force between ordinary objects (like two people) is negligible/unnoticeable, but it is significant for massive bodies (like Earth and Sun).

According to Newton's Third Law, action and reaction are equal but opposite. Similarly, the gravitational force of mass $m_1$ on $m_2$ is equal and opposite to the gravitational force of mass $m_2$ on $m_1$. They form an action-reaction pair.

Consider a body of mass $m$ on the surface of the Earth. Let Earth's mass be $M_e$ and radius be $R$.
The gravitational force between body and Earth is:
$F = G \frac{m M_e}{R^2}$
This force is also the weight of the body ($W = mg$). Therefore:
$mg = G \frac{m M_e}{R^2}$
Canceling $m$:
$g = G \frac{M_e}{R^2}$
Rearranging for $M_e$:
$M_e = \frac{R^2 g}{G}$

Using standard values:
Radius of Earth ($R$) = $6.4 \times 10^6 m$
Gravitational Acceleration ($g$) = $10 ms^{-2}$
Gravitational Constant ($G$) = $6.673 \times 10^{-11} Nm^2kg^{-2}$

Substituting these:
$M_e = \frac{(6.4 \times 10^6)^2 \times 10}{6.673 \times 10^{-11}}$
$M_e \approx 6.0 \times 10^{24} kg$

The value of gravitational acceleration '$g$' depends on the distance from the center of the Earth. The formula for '$g$' at an altitude '$h$' is:
$g_h = \frac{G M_e}{(R+h)^2}$
This shows that '$g$' is inversely proportional to the square of the distance from the Earth's center. As altitude ($h$) increases, the value of '$g$' decreases.

At a height equal to one Earth radius ($h=R$), the distance from the center becomes $2R$.
$g_h = \frac{G M_e}{(2R)^2} = \frac{1}{4} \frac{G M_e}{R^2} = \frac{1}{4} g$
So, at height $R$, the value of $g$ becomes one-fourth of its value on the surface.

Satellite: An object that revolves around a planet is called a satellite.
- Natural Satellite: The Moon revolves around the Earth, so it is a natural satellite.
- Artificial Satellites: Man-made objects sent into space to orbit the Earth are called artificial satellites. They are used for communication, weather forecasting, and navigation.

The velocity required by a satellite to keep moving in its orbit is called orbital velocity ($v_o$).
For a satellite revolving close to the Earth, the orbital velocity is given by:
$v_o = \sqrt{gR}$
Putting values ($g=10 ms^{-2}, R=6.4 \times 10^6 m$):
$v_o \approx 8000 ms^{-1} = 8 kms^{-1}$
This is roughly $29,000 kmh^{-1}$.

Satellites whose period of revolution is exactly 24 hours (same as Earth's rotation) appear stationary from Earth. These are called Geostationary Satellites. They are used for global communication and orbit at a height of approx $42,300 km$ from the center of Earth.