The gravitational potential energy U(r) of a system of two masses m₁ and m₂ separated by a distance r as the amount of work done to bring the mass m₂ from infinity to a distance r assuming m₁ to be fixed in its position and is written as
\(\mathrm{U}(\mathrm{r})=-\frac{G m_{1} m_{2}}{r}\)

Consider 'n' particles of masses \(m_{1}, m_{2}, \ldots \ldots m_{n}\), distributed in space at positions \(\overrightarrow{r_{1}}, \overrightarrow{r_{2}}, \overrightarrow{r_{3}}, \ldots .\) etc, with respect to point P. The total gravitational field at a point P due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of gravitational fields.
\(\vec{E}_{\text {total }} =\vec{E}_{1}+\vec{E}_{2}+\ldots \vec{E}_{n}  \)
\(=-\frac{G m_{1}}{r_{1}^{2}} \hat{r}_{1}-\frac{G m_{2}}{r_{2}^{2}} \hat{r}_{2}-\ldots-\frac{G m_{n}}{r_{n}^{2}} \hat{r}_{n}=-\sum_{i=1}^{n} \frac{G m_{i}}{r_{i}^{2}} \hat{r}_{i}\)

The planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called "retrograde motion" of planets.
To explain this retrograde motion, Ptolemy introduced the concept of "epicycle" in his geocentric model. According to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as "epicycle". A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth.

Because two masses come together slowly from infinity, work is done by the system.

This model successfully explained the motion of all celestial objects. But the motion of Mars and Jupiter could not be explained effectively.

Nicholas Copernicus (1473-1543) proposed a new model called the 'Heliocentric model' in which the Sun was considered to be at the center of the solar system and all planets including the Earth orbited the Sun in circular orbits.

Tides arise in the ocean due to the force of attraction between the moon & sea water.

(i) The law of conservation of angular momentum.
(ii) Yes,∵ Kepler's III law is. the relation between distance & time.

The acceleration due to gravity decreases due to rotation of the earth. This effect is zero at poles & maximum at the equator.

The value of g is less on hills because they are comparatively at a greater distance from the centre of the earth. ஃ The gravitational pull on the tennis bell is less on hill tops & so it bounces higher on hills than on plains.

\(\mathrm{g}_{\text {latitude'}} \ g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda\)
Value of latitude of g at Chennai \(\simeq 13^{\circ}\)
\(\operatorname{Cos} 13^{\circ} =0.2268 \mathrm{rad}  \)
\(\omega =\frac{2 \pi}{T}=\frac{2 \pi}{24 \times 3600}  \)
\(=\frac{2 \pi}{86400}=\frac{2 \times 3.14}{86400}  \)
\(\therefore \omega^{2} R =\left(\frac{2 \times 3.14}{86400}\right)^{2} \times\left(6400 \times 10^{3}\right)  \)
\(=3.4 \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2} \)
\(\mathrm{g}_{\text {Cbeanai }} =\mathrm{g}-\omega^{2} R \cos ^{2} \lambda  \)
\(=9.8-\left(3.4 \times 10^{-2}\right)^{2} \cos (0.2268)^{2}  \)
\(\mathrm{~g}_{\text {Chennai }} =9.7677 \mathrm{~m} / \mathrm{s}^{2}\)
Value of latitude at Kanyakumari
\(=8.088^{\circ} \mathrm{N}=8.08=8.1^{\circ} \mathrm{N}  \)
\(\omega =\frac{2 \pi}{T}=\frac{2 \times 3.14}{86400}  \)
\(\omega^{2} R =\left(\frac{2 \times 3.14}{86400}\right)^{2} \times\left(6400 \times 10^{3}\right)  \)
\(=3.4 \times 10^{2} \mathrm{~m} / \mathrm{s}^{2}\)
\(\mathrm{g}_{\text {Kanyakumari }} =g-\omega^{2} R \cos ^{2} \lambda \)
\(=9.8-\left(3.4 \times 10^{2}\right)^{2}\left[\cos \left(8.1^{\circ}\right)\right]^{2} \)
\(g_{\text {Kanyakumari }} =9.798 \mathrm{~ms}^{-2} \)
\(\Delta g =9.798-9.767=0.031 \mathrm{~ms}^{-2}\)

Moon will start to move in spiral path towards the earth and it may hit on the surface of the Earth.

Kinetic energy of the moon will vary but its total energy remains constant.

The gravitational potential at a distance r due to a mass is defined as the amount of work required to bring unit mass from infinity to the distance r and it is denoted as V(r).
\(\mathrm{V}(\mathrm{r})=-\frac{\mathrm{Gm}}{r}\)