
What is the importance of the universal law of gravitation?
Answer
590.1k+ views
Hint: Try to form an equation for the gravitational force, then infer its dependency on different parameters.
Formulae used: Essentially, the formula for the gravitational force between two objects is used. The formula is,
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
Complete step-by-step answer:
The universal law of gravitation or Newton's law of gravitation states that any object of mass in the universe attracts any other with a force varying directly as the product of the masses of the objects and inversely as the square of the distance between them.
The important observations deduced from the law are as follows:
1. The gravitational force or gravity between two masses becomes stronger with increasing the mass of an object.
2. Similarly, if the masses of the two objects are fixed, the gravity between them weakens if they are further distanced, as the distance between them increases.
- This law has been successful in explaining many phenomena especially concerning celestial bodies such as:
Revolution of planets and moons around the sun and the planets respectively.
Existence of gravity, an attractive force, on Earth.
Origination of tidal waves.
And many more…
Additional Information:
Now, we will try to write an equation from the above statement. The law shows that the gravitational force only depends on the masses of the objects and the distance between them. So, let the mass of the first object be ${{m}_{1}}$ , mass of the second object be ${{m}_{2}}$ and the distance between them be $d$ .
Now, according to the first sentence of the law, the gravitational force, let us take $F$ , is proportional to the product of the mass of the objects. Therefore, in mathematical terms it can be written as
$F\propto {{m}_{1}}{{m}_{2}}$
Second line states that the force is inversely proportional to the square of the distance between the two objects. Which gives,
$F\propto \dfrac{1}{{{d}^{2}}}$
Combining the two proportionalities, we get
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
When the experiments were performed to verify this law, it was found that to get the accurate value of the gravitational force, a proportionality constant had to be multiplied. The constant was then named the Gravitational constant $G$ . Its value is
$G=6.673\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Therefore, the final equation for gravitational force is
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
Note: The law assumes the objects to be point sized. When considering celestial objects, since the distances are very large and the masses are huge, the bodies are approximated as point sized. All the theories proposed for the celestial bodies are some assumed approximations.
Formulae used: Essentially, the formula for the gravitational force between two objects is used. The formula is,
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
Complete step-by-step answer:
The universal law of gravitation or Newton's law of gravitation states that any object of mass in the universe attracts any other with a force varying directly as the product of the masses of the objects and inversely as the square of the distance between them.
The important observations deduced from the law are as follows:
1. The gravitational force or gravity between two masses becomes stronger with increasing the mass of an object.
2. Similarly, if the masses of the two objects are fixed, the gravity between them weakens if they are further distanced, as the distance between them increases.
- This law has been successful in explaining many phenomena especially concerning celestial bodies such as:
Revolution of planets and moons around the sun and the planets respectively.
Existence of gravity, an attractive force, on Earth.
Origination of tidal waves.
And many more…
Additional Information:
Now, we will try to write an equation from the above statement. The law shows that the gravitational force only depends on the masses of the objects and the distance between them. So, let the mass of the first object be ${{m}_{1}}$ , mass of the second object be ${{m}_{2}}$ and the distance between them be $d$ .
Now, according to the first sentence of the law, the gravitational force, let us take $F$ , is proportional to the product of the mass of the objects. Therefore, in mathematical terms it can be written as
$F\propto {{m}_{1}}{{m}_{2}}$
Second line states that the force is inversely proportional to the square of the distance between the two objects. Which gives,
$F\propto \dfrac{1}{{{d}^{2}}}$
Combining the two proportionalities, we get
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
When the experiments were performed to verify this law, it was found that to get the accurate value of the gravitational force, a proportionality constant had to be multiplied. The constant was then named the Gravitational constant $G$ . Its value is
$G=6.673\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Therefore, the final equation for gravitational force is
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$
Note: The law assumes the objects to be point sized. When considering celestial objects, since the distances are very large and the masses are huge, the bodies are approximated as point sized. All the theories proposed for the celestial bodies are some assumed approximations.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

