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Universal Law of Gravitation

Universal Law of Gravitation

Table of Content

Introduction of Gravitation

Gravity is one kind of attraction that drags us toward Earth but it's also that attraction which drags Earth toward us. This is simplified by the universal law of gravitation, which explains how all objects in the universe have this kind of force among them. Most of us are aware Newton’s conclusion on an apple falling from a tree and subsequent discovery of gravitational force.
Gravitation

The Force of Gravity

Assume, there is a force between two masses. The net force on you decides how your motion will modify. The force of gravity between you and the desk is extremely small compared with other forces. It will always act on you, such as friction, the force which is from your muscles, Earth’s Gravity, and the gravitational pull from many other objects. The durability of the gravitational force between two objects depends on two factors, mass, and distance. The mass of the two objects that means the more mass two objects have, the greater the force of gravity the masses apply on each other. If one of the masses is doubled, the force of gravity between the objects is doubled.
The force of Gravity

What is Centripetal Force?

Newton showed that the force responsible for the fall of that apple was the same as the force that held at the moon in its orbit. This implies that the force of attraction between any two bodies applies to all the bodies in the universe. This force is referred to as gravitational force. The word gravity is derived from Latin meaning “Heavy”. The moon revolving around the earth is under the action of the centripetal force(F) which makes it follow a curved path.

Centripetal Force

This centripetal force is provided by the gravitational force between the earth and the moon. Similar to that of the force between the moon and the earth. The immensity of the gravitational force between any two objects varies with the distance between them as well as their masses. According to Newton the moon revolving around the earth as a centripetal force which is due to the earth gravitational force.

Derivation

The Centripetal acceleration of the moon is given by the expression:

Where V = speed of the moon in its orbit.

R= Distance between the moon and the earth

The velocity of the moon in its orbit is derived from the fact that the moon orbit is circular of radius Rm and its revolution period is t.Thus the orbital velocity of moon,

Where, T = Period of revolution of the moon

From equation (1), we will get,

Radius of the moon’s orbit, R= 3.8 × 108m

Time period of moon, T = 27.3 days ≅ 2.36 × 106 s

Substituting the value of Rm and T in the expression and simplifying

The acceleration of the moon obtained is much less than the acceleration of a freely falling body of the surface of the earth.That means,

From the observation of equation (2), Newton concluded;

The Universal Law of Gravitation

The magnitude of the earth’s gravitational force decreases with distance. Thus based on the observation regarding forces, Newton proposed The Universal Law of Gravitation. It states that “Every particle in the universe attracts other particles with a force that is:

  • Directly proportional to the product of their masses
  • Inversely proportional to the square of the distance between them”.

The Universal Law of GravitaitonThus the law can be expressed as:

Combining the two expressions, we get

Here m1 and m2 point masses and r are the distance between two particles. Replacing the proportionality sign with the equal to a symbol, we got a constant G.

Where G = Universal gravitational constant and it’s valued.

G = 6.67 × 10-11 Nm2 kg-2

Throughout the cosmos, there are some latent principles that we assume to be steady; as an example, the speed of light in a vacuum. So many different facts of our understanding of the universe come from this one constant. If the speed of light wasn’t steady, Global Positioning Systems wouldn’t work, and we wouldn’t be able to determine the age of far galaxies in the universe. Most of the Einstein’s equations would drop apart without a constant speed of light; time expansion and, the mass-energy relationship would vary and quantum changes on the Planck scale could probably have large scale effects. The speed of light in a vacuum is considered to be constant, never changing over the last 13 billion years.Yet, there are more than other constants, and the one we are going to be discussing, which is the gravitational constant.

The Gravitational Constant

The Gravitational Constant value is 6.67384 ×10-11m3 kg-1s-2. Now, the possibility looks a bit complex but basically it means that gravity has a set of strengths. A strength that works in three-dimensional space, is an accelerating force which is determined by mass. The reason is the set value of the gravitational constant that we’re able to accurately determine the mass of two orbiting objects. By calculating the amount of force gave on us by the Earth to keep us on the Earth which is 9.81 m/s2. So far, the gravitational constant has remained constant throughout the entire history of the universe. Calculation of the gravitational constant over the past 200 years have been changeable. Even as the methods that we use now are far more advanced and sensitive than were used two centuries ago, the true value of the gravitational constant is not easy to find.

The Gravitational Constant
In 2013, Some researchers made a team and working out of France took the calculation of the gravitational constant, which is using the same machine that they used some two years earlier. Some improvements made on the machine to improve the sensitivity and give more precise result. The machine, which are using two independent methods to calculate the constant, averages the results of the two. This, in theory, should help to close the systematic mistake.

Mathematical Expression

Now let’s consider a coordinate system having position vector rand r2, when we apply the law two point masses mand min the coordinate system.

Coordinate SystemWe get the mathematical expression, for the gravitational force mon mas F12

Where r12 = the vector in the direction mto m2.

=r2 – r1

Here the negative sign indicates the force is an attractive force and the direction of minus r12, which is mto m1.

Mathemmatical ExpressionThus the same expression can be written as:

Where hat{r} = Unit vector in the direction m1to m2. In this expression

If F21 represent the gravitational force on point mass mdue to point mass m2. Then F12 represent the gravitational force on mdue to m1. According to Newton’s third law of motion, both the forces,

Now, let’s consider a system with more than two point masses m1, m2, m3, m4, m5 placed at a different position in space. The net gravitational pull forces exerted on a single mass say mis the vector sum of the gravitational pull forces of all the other four masses on m1. Thus net gravitational force on the particle of m1.

A system with more than two point massesThus net gravitational force on the particle of mass mis given by F1,net

Gravitational forces between extended objects

Now let’s try to find gravitational forces between extended objects which are huge in sizes like the earth and the moon. For this, we need to find the vector sum of the gravitational forces exerted by all the particles that constitute the body. This can be obtained by using calculus. When we deal with the gravitational forces between extended bodies, we need to consider two special cases.

The first one/case 1

The gravitational force of attraction between the hemispherical shell of uniform mass, density and a point mass resting outside of the surface is equal to the gravitational force between the point mass and the mass of the shell concentrated its center.

The first one case 1

The second one/case 2

If we consider a point mass within a hemispherical uniform density, the force of attraction due to the shell on a point mass situated inside it is zero.

The Second one Case
Both case in details: Case 1

The hemispherical shell can be considered to be consisting of several point masses along its surface. For Example, consider the circular strip of the shell and the point masses on it. Gravitational forces caused by the various point of masses on the circular strip of the shell have two components of the forces, one is the line joining point mass and the center of the shell.

Both case in details Case 1
And the other perpendicular to it. The perpendicular components cancel each other, making the resultant force acting along the line joining the point mass and the center of the shell. The magnitude of this net force is equal to the force between the point mass outside the shell and the total mass of the shell concentrated at its center.

Both case in detials
Case 2:

In this case, again consider different regions of the spherical shell that contain point masses. The point masses on a different region of the shell attract the point mass inside it in different directions. These forces cancel each other completely and hence the resultant gravitational force on point mass inside it zero.
Case Two

Thus, the gravitational force between extended objects differs from one situation to another.

Frequently Asked Questions (FAQs)


Q1
. Is Gravity is a force that can be changed?

Sol. Exactly, it’s true. Gravity always unchangeable across the surface of the Earth and throughout its atmosphere, due to some effects.

First one, there is the difference of gravity with latitude that already mentioned to you weigh about 0.5% more at the poles than on the equator. There are two outcomes that contribute to this. It should be noted, yet, one of these results is due to an actual difference in the gravitational force between the equator and poles. The other effect is due to the fact that the Earth is revolving, which affects the weight you would see when you stepped on a scale but does not actually represent a vary in the value of the gravitational force.

Secondly, gravity never change with altitude. The gravitational force above the Earth's surface is proportional to 1/R2, where R is the distance from the center of the Earth. The radius of the Earth at the equator is 6,378 kilometers, so assume you were on a hill at the equator that was 5 kilometers high (around 16,400 feet). You would then be 6,383 kilometers from the Earth's center, and the gravitational force would have decreased by a factor of (6,378 / 6,383)2 = 0.9984. So the difference is less than 0.2%.

So, there are very small differences on the order of 0.01% or less in gravity due to differences in the local geology. As an example, varies in the density of rock underneath you or the presence of mountains nearby can have a tiny effect on the gravitational force.

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