Gravitational Force Escape Velocity - Detailed Guide

What is Gravitational Force

Gravitational force definition: Gravitational force means the force of attraction between any two objects. The tendency of particles to move towards each other is called gravitation. The greater the mass of an object, the stronger the gravitational pull. Gravitational force depends on the :

• Masses of the objects
• The distance between them

Formula of Gravitational Force

The formula of gravitational force is given as:

$$
F=\frac{G M_1 M_2}{R^2}
$$

where,

• $F$ is the gravitational force between the two objects
• $G$ is the gravitational constant
• $M_1$ and $M_2$ are the masses of the two objects
• $R$ is the distance between the centers of the two objects

Gravitational Force Examples

What is gravitational force? Have you ever wondered why anything thrown up falls? Why doesn’t the moon fall to Earth? Why doesn’t the Earth fly off into the vast space rather than rotating in its orbit around the sun? Why do all the planets revolve around the sun? Why is going upstairs a lot more tiring than going downstairs? In all these cases there must be some force acting on the moon, the planets, and the falling bodies the force acting on them is called gravitational force one of the standing goals of physics is to understand the gravitational force. The force that holds us to the earth, holds all the objects surrounding you to earth, holds the moon in orbit revolving around the earth holds the earth and all planets in their path around the sun. This force is the reason for our Milky Way galaxy holding together billions of stars in the universe and the countless astronomical bodies between stars.

Gravitational Force Discovery

Gravitational force is a type of non-contact force. This is the force when the two interacting bodies are not in direct physical contact with each other, here, they can exert a push or pull despite their separation.

So, the question here is what causes the gravitational force?

There is a popular story that one day, Newton was sitting under an apple tree when an apple fell on his head and he suddenly thought of the force of gravitation. He thought about the following lines-

If the apple is accelerated there must be a force acting on it. If the force is at the top of the highest level the tree might not reach even further to the moon. By such reasonings, Newton concluded that,

Any two objects in the universe exert gravitational force of attraction on each other and he proposed the definition of gravitational force, as the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Till now, we have understood the meaning of gravitational force very clearly and the importance of gravitational force on Earth as well as in the whole universe.

Calculate Acceleration Due to Gravity

According to Newton’s universal law of gravitational force,

$F=\frac{G M_1 M_2}{R^2} \ldots$ (1)

Where, $G=6.673 \times 10^{-11}$ (Gravitational Constant)
$M_1=1.99 \times 10^{30} \mathrm{~kg}$ (Mass of the sun in this case)
$R=6.96 \times 10^8 \mathrm{~m}$ (Radius of the sun)

As the force on a body is given as,

$
F=M_2 g
$.... (2)

From equation (1):

$$
F=\frac{G M_1 M_2}{R^2}
$$

Equating equations (1) and (2):

$$
M_2 g=\frac{G M_1 M_2}{R^2}
$$

Dividing both sides by $M_2$ :

$$
g=\frac{G M_1}{R^2}
$$

By substituting all the values, we get the acceleration due to gravity as:

$$
g=9.8 \mathrm{~m} / \mathrm{s}^2
$$

Calculate the Gravitational Pull Of Earth

As the moon revolves around the earth moon experiences a centripetal acceleration directed toward the earth an object near the earth's surface such as the apple experiences an acceleration g.

Newton calculated the acceleration of a body toward the earth is inversely proportional to the square distance of the body from the center of the earth by using the inverse square law.

According to the formula of inverse square law-

$$\frac{g}{A_M}=\frac{\left(\frac{1}{R_E}\right)^2}{\left(\frac{1}{R_M}\right)^2}$$

Where,

• $g$ is the acceleration due to gravity
• $R_E$ is the radius of the Earth
• $A_M$ is the centripetal acceleration
• $R_E$ is the distance between the Earth and the Moon

By substituting the values,

$$
R_M=3.85 \times 10^8 \mathrm{~m} \text { and } R_E=6.38 \times 10^6 \mathrm{~m}
$$

We get,

$$
\frac{g}{A_M}=3600 \text { (approx.) }
$$

And,

$$
\begin{gathered}
A_M=\frac{9.8}{3600} \\
A_M=2.7 \times 10^{-3} \mathrm{~m} / \mathrm{s}^2
\end{gathered}
$$

Also,

$$
\frac{R_M}{R_E}=60
$$

Therefore,

$$
g=9.8 \mathrm{~m} / \mathrm{s}^2
$$

Newton also calculated the centripetal acceleration of the moon from a knowledge of its mean distance from the Earth and its orbital.

As Centripetal Acceleration is $A_M=\frac{V^2}{R_M}=\frac{4 \pi^2 R_M}{T^2}$
Where, $V=\frac{2 \pi R_M}{T}$
By substituting the value of $T$ as 27.3 days and $R_M=3.85 \times 10^8 \mathrm{~m}$,
We get, $A_M=2.7 \times 10^{-3} \mathrm{~m} / \mathrm{s}^2$
Similarly, the gravitational force of the Sun is also calculated.
The gravitational pull between the Earth and the Sun is equal to $3.52 \times 10^{22}$ newtons.

The gravitational force on different planets is different because of their varying mass. More is the mass of the planet more is the gravitational force applied by it. Also closer anybody is more the gravitational force. Hence, in our solar system Jupiter has the maximum force on Earth and 2nd planet is Venus and Mercury has the lowest gravitational pull.

What is Escape Velocity

It is the minimum velocity required for an object to escape the gravitational pull of a massive celestial body. The escape velocity depends on the mass of the heavy body and the distance from its center to the object. The escape velocity of Earth is 11.2 km/s

Escape Velocity Formula

The formula of escape velocity can be expressed as:

$$
v_e=\sqrt{\frac{2 G M}{R}}
$$

where,

• $G$ is the gravitational constant
• $M$ is the mass of the celestial body
• $R$ is the radius of the celestial body

Derivation Of Escape Velocity

The gravitational potential energy of an object from the center of a celestial body is given as

$U=-\frac{G M m}{r}$

The kinetic energy of the object is

$K=\frac{1}{2} m v^2$

To escape the escape velocity total energy of the object must be zero or greater

Thus,

$\frac{1}{2} m v^2-\frac{G M m}{r} \geq 0$

$\frac{1}{2} m v^2=\frac{G M m}{r}$

$v^2=\frac{2 G M}{r}$

Taking the square root we get

$$v_{\text {escape }}=\sqrt{\frac{2 G M}{r}}$$

Factors Affecting Escape Velocity

1. Mass of the celestial body
2. Radius of the celestial body
3. Gravitational Constant
4. External gravitational influence

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