# Chapter 4: Gravitation 🪐

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## **Introduction**

[cite_start]Physics has long sought to answer fundamental questions about the cosmos, such as why the moon doesn't fall to Earth and why planets remain in orbit around the sun[cite: 3, 4]. [cite_start]The study of **gravitation** provides the answers to these questions[cite: 5]. [cite_start]Gravitation is the tendency of bodies to move towards one another[cite: 9]. [cite_start]Of the four fundamental forces in nature, gravitation was one of the first to be studied extensively[cite: 6]. [cite_start]In the 17th century, Sir Isaac Newton discovered that the same force that makes an apple fall from a tree is the one that holds the moon in its orbit[cite: 7]. [cite_start]He concluded that every body in the universe attracts every other body[cite: 8].

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## **Newton's Law of Gravitation**

[cite_start]This law states that "everybody in the universe attracts each other with force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers." [cite: 11]


[cite_start]Consider two bodies of masses $m_1$ and $m_2$ separated by a distance 'r'[cite: 12]. The force (F) between them follows these relationships:
* [cite_start]The force is directly proportional to the product of their masses[cite: 15]:
    $$ F \propto m_1m_2 $$
* [cite_start]The force is inversely proportional to the square of the distance between them[cite: 16]:
    $$ F \propto \frac{1}{r^2} $$

[cite_start]Combining these, we get[cite: 21]:
$$ F \propto \frac{m_1m_2}{r^2} $$
[cite_start]This can be written as an equation with a proportionality constant, G[cite: 24, 26]:
$$ F = \frac{Gm_1m_2}{r^2} $$
[cite_start]The direction of this gravitational force is along the line that joins the centers of the two bodies[cite: 27].

### **Universal Gravitational Constant (G)**
[cite_start]'G' is known as the **universal gravitational constant**[cite: 26]. We can define it by considering a special case where the masses and distance are unity. [cite_start]If $m_1=m_2=1$ and $r=1$, then the equation becomes $F=G$[cite: 28].

[cite_start]Therefore, the universal gravitational constant G is numerically equal to the force of attraction between two bodies of unit mass separated by a unit distance[cite: 29].
* [cite_start]**SI Unit**: $Nm^2kg^{-2}$[cite: 30].
* [cite_start]**Value**: $6.67 \times 10^{-11} Nm^2kg^{-2}$[cite: 30].

#### **Properties of G**
* [cite_start]It is independent of the nature of the masses of the body[cite: 32].
* [cite_start]It is unaffected by temperature and pressure[cite: 33].
* [cite_start]It is independent of the directions of the force between the two masses[cite: 34].

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## **Gravitational Field and Intensity**

### **Gravitational Field**
[cite_start]The gravitational field of a mass is the space around it where another mass will experience its gravitational force[cite: 36].

### **Gravitational Field Intensity (E)**
[cite_start]The gravitational field intensity at a point is defined as the force experienced by a unit test mass placed at that point[cite: 37].
* [cite_start]It is a measure of the strength of the gravitational field[cite: 38].
* [cite_start]It is a **vector quantity**, and its direction is towards the body producing the field[cite: 38].
* [cite_start]It is denoted by **E**, and its unit is **N/kg**[cite: 39].

#### **Expression for Gravitational Field Intensity**
[cite_start]Consider a large mass M (like the Earth) and a small test mass 'm' placed at a point P, which is a distance 'r' from the center of M[cite: 40, 41].
* [cite_start]The gravitational force on the test mass 'm' is given by Newton's law[cite: 41]:
    $$ F = \frac{GMm}{r^2} $$
* [cite_start]By definition, the intensity E is the force per unit mass[cite: 51]:
    $$ E = \frac{F}{m} = \frac{GMm/r^2}{m} $$
* [cite_start]This simplifies to the formula for gravitational field intensity[cite: 50]:
    $$ E = \frac{GM}{r^2} $$
[cite_start]From this equation, we can see that the gravitational field intensity at a point is equal to the acceleration produced on a point mass at that point due to gravity[cite: 54, 55].

* [cite_start]**On the surface of the Earth**: If the point is on the surface, then $r=R$ (the radius of the Earth)[cite: 56]. [cite_start]The intensity becomes[cite: 58]:
    $$ E = \frac{GM}{R^2} = g = 9.8 \text{ N/kg} $$

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## **Gravitational Potential and Potential Energy**

### **Gravitational Potential Energy (U)**
[cite_start]The gravitational potential energy at a point in a gravitational field is defined as the amount of work done to bring a body from infinity to that point[cite: 61]. [cite_start]It is a **scalar quantity** and is denoted by U[cite: 62].

### **Gravitational Potential (V)**
[cite_start]The gravitational potential at a point is the work done in bringing a **unit mass** from infinity to that point without acceleration[cite: 63]. [cite_start]It is simply the gravitational potential energy per unit mass[cite: 64].
$$ V = \frac{U}{m} $$

#### **Expression for Gravitational Potential Energy**
[cite_start]Let's derive the formula for the potential energy of a body of mass 'm' at a point C, which is a distance 'r' from the center of the Earth (mass M)[cite: 68, 80].
1.  [cite_start]Imagine moving the body from infinity towards point C. Consider an intermediate point A on this path, at a distance 'x' from the Earth's center[cite: 81].
2.  [cite_start]The gravitational force on the body at point A is[cite: 82, 83]:
    $$ F = \frac{GMm}{x^2} $$
3.  [cite_start]Now, let's find the small amount of work done (dW) in moving the body a tiny distance 'dx' from point A to point B[cite: 85].
    $$ dW = Fdx = \frac{GMm}{x^2}dx $$
4.  [cite_start]To find the total work done (W) in bringing the body all the way from infinity to point C (a distance 'r'), we integrate dW from the starting point ($x=\infty$) to the final point ($x=r$)[cite: 88].
    $$ W = \int_{\infty}^{r} dW = \int_{\infty}^{r} \frac{GMm}{x^2}dx $$
5.  Solving the integral:
    $$ W = GMm \int_{\infty}^{r} x^{-2}dx = GMm \left[ \frac{x^{-1}}{-1} \right]_{\infty}^{r} = GMm \left[ -\frac{1}{x} \right]_{\infty}^{r} $$
6.  Now, we apply the limits of integration:
    $$ W = GMm \left[ \left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right) \right] = GMm \left[ -\frac{1}{r} + 0 \right] $$
7.  [cite_start]This work done is stored as gravitational potential energy (U)[cite: 90].
    $$ U = W = -\frac{GMm}{r} $$