**Newton's Law of Gravitation**

Gravity is the phenomenon of attraction of all bodies in the world which is described by the Newton's Law of Universal Gravitation. This law states that:

*"Two material points with masses \(m_1\) and \(m_2\) placed at a distance \(r\) from each other are attracted with the force proportional to their masses and inversely proportional to the square of the distance between them"*:

\(F=G\frac{m_1m_2}{r^2}\), where \(G=6,67 \cdot 10^{-11} \frac{N\cdot m^2 }{kg^2} \)called as the gravitational constant.

This formula is also valid for homogeneous spheres and balls

(r is the distance between the centers of the bodies).

The gravitational pulling forces act along the line connecting the centers of the bodies, and according to the Third Newton's law of motion are equal in magnitude and opposite in direction:

\(\vec{F_1}=-\vec{F_2}\)

**Weight** is a force of attraction of the body of mass \(m\) by the Earth

\(W=G\frac{Mm}{r^2}=mg_h,\)

where \(M\) is the mass of Earth; \(r=R+h\) is the distance from the body to the center of the Earth; \(R\) is the Earth's mean radius and \(h\) is the distance from the body to the Earth's surface; \(g_h\)is the gravitational acceleration at height \(h\) above the Earth's surface.

The Earth's gravitational force has direction to the center of the Earth.

The gravitational acceleration at height \(h\) above the Earth's surface is

\(g_h=G\frac{M}{r^2}=G\frac{M}{(R+h)^2}\)

On the surface of the Earth gravitational acceleration equals \(g=G\frac{M}{R^2}=9.8\frac{m}{s^2}.\)

The body that revolves around the planet or star due to gravity is called **satellite**. The plane of the satellite's orbit passes through the center of the planet or star.

**Orbital velocity** \(v_0\) is the speed that should be given to a body at a distance \(r=R+h\) from the center of planet or star, to make it move in the circular orbit, whose center coincides with the center of the planet or star. By Newton's second law \(ma_c=F_g=G\frac{Mm}{r^2}.\) Centripetal acceleration equals \(a_c=\frac{v^2}{r}.\)Then the orbital velocity is

\(v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+h}}.\)

If the satellite is at an altitude \(h \ll R\) then \(v_o=\sqrt{\frac{GM}{R}}=\sqrt{gR}.\) On the surface of the Earth \(v_o=7.9 \frac{km}{s}.\)

**Escape velocity** is the minimum speed needed for an object to escape from a planet: \(v_e=v_o\sqrt{2}\). On the surface of the Earth, the escape velocity is \(v_e=11.2 \frac{km}{s}\).

**Example 1.**

Satellite moves in circular orbit at an altitude of 6400 km above the Earth. Determine the speed of the satellite in this orbit. The radius of the Earth equals 6400km.

**Solution:**

The velocity of the satellite in circular orbit whose radius \(r=2R\) can be calculated by the formula \(v=\sqrt{\frac{GM}{2R}}\). Since the gravitational acceleration on Earth is \(g=\sqrt{\frac{GM}{R}}\), then \(GM=gR^2\). Finally,

\(v=\sqrt{\frac{gR^2}{2R}}=\sqrt{\frac{gR}{2}}=5.6 \frac{km}{s}.\)

**Example 2.**

A satellite appears motionless to ground observer located on the equator. Determine the radius \(r\) of the orbit of the satellite. The radius of the Earth is \(R=6400\; km.\)

**Solution:**

Orbital period of the motionless to ground observer satellite coincides with the rotation period of the Earth, namely \(T_o=24h.\) This orbital period is connected with centripetal acceleration of the satellite by

\(a=\omega^2r=\left(\frac{2\pi}{T_o}\right)^2r=\frac{4\pi^2r}{T_o^2},\)

with \(\omega\) being the angular velocity of the satellite.

On the other hand, centripetal acceleration \(a\) can be found from the Newton's second law \(a=\frac{F_g}{m}=\frac{GM}{r^2}.\)

From these expressions for centripetal acceleration we obtain \(r=\sqrt[3]{\frac{GMT_o^2}{4\pi^2}}.\)

Remembering that \(GM=gR^2\), we finally write

\(r=\sqrt[3]{\frac{gR^2T_o^2}{4\pi^2}}=42.6\cdot10^6 m.\)