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The gravitational constant, denoted by G is a physical constant that is involved in the formula for calculating gravitational force between two bodies.
- In Newton’s universal law of Gravitation, it is the proportionality constant that relates the gravitational force between two bodies to the product of their masses and the inverse of the square of the distance between them.
- The value of the gravitational constant was first measured experimentally by the English scientist Henry Cavendish in 1798.
- His experiment for the measurement of the gravitational constant is famously known as the Cavendish experiment.
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Key Terms: Universal law of Gravitation, Force, Gravitational force, Kepler’s Laws, Cavendish experiment, SI units.
Universal Law of Gravitation
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Sir Issac Newton proposed that all particles or objects in the universe attract each other in the same manner as the earth attracts the apple. The force of attraction between any two bodies of this universe is called Gravitation or Gravitational Force.
“According to the Universal Law of Gravitation, every object in the universe attracts every other object with a force that is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them.”
Mathematically,
F ∝ \(\frac{m_1m_2}{r^2}\)
Where
- F is the gravitational force
- m1 and m2 are the masses of the two objects
- r is the distance between the bodies
On removing the proportionality sign, we get
F = G \(\frac{m_1m_2}{r^2}\)
Where G is a constant of proportionality, known as Gravitational constant.
The video below explains this:
Relation Between G And g Detailed Video Explanation:
Also Read: Value of g on Moon
Gravitational Constant
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The constant of proportionality G obtained in the equation of the universal law of gravitation, is known as Gravitational constant. The measured value in the SI unit of this physical constant is
G = 6.674 x 10-11 m3 kg-1 s-2
Or G = 6.674 x 10-11 N m2 kg-2
The gravitational constant is also known as
- Universal Gravitational Constant
- Newtonian Constant of Gravitation
- Cavendish Gravitational Constant
G = 6.674 x 10-11 N m2 kg-2
Value of gravitational constant
The dimensional formula of Gravitational constant is [ M-1 L3 T-2 ].
A few important points about the gravitational constant are
- The value of gravitational constant G was calculated by Henry Cavendish in 1798.
- Gravitational constant is a universal constant that remains constant throughout the universe.
- It is independent of the nature of the two objects (i.e. size, shape, mass, etc.) at all places and at all times throughout the universe.
- In experiments for determining the value of gravitational highly sensitive and precise equipment are required.
- The gravitational constant is used in the equations of the Law of Gravitation, Einstein’s General Theory of Relativity, and also Kepler’s Third Law of Planetary Motion to calculate the time period of a planet.
Also Read:
Things to Remember
- Gravitational constant is a physical constant involved in the formula for calculating gravitational force between two bodies.
- The value of the gravitational constant is G = 6.674 x 10-11 m3 kg-1 s-2
- Henry Cavendish in 1798 first measured experimentally the value of the gravitational constant.
- According to the Universal Law of Gravitation, the force of attraction between two objects is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them.
Sample Questions
Ques. What is the value of the gravitational constant in the SI unit? (1 Mark)
Ans. In the SI unit, the value of the gravitational constant is G = 6.674 x 10-11 m3 kg-1 s-2
Ques. A sphere of 20 kg is attracted by a second sphere of mass of 50 kg with a force equal to 0.01 gf. If G = 6.67 x 10-11 N m2 kg-2, calculate the distance between them. (3 Marks)
Ans. Given
- Mass of the first sphere, m1 = 20 kg
- Mass of the second sphere, m2 = 50 kg
- Gravitational force between the spheres, F = 0.01 gf = 9.8 x 10-5 N
The formula for gravitational force between two bodies is given by
F = G \(\frac{m_1m_2}{r^2}\)
Where r is the distance between the two bodies.
Therefore,
r = \(\sqrt{G\frac{m_1m_2}{F}}\) = \(\sqrt{\frac{6.67 \times 10^{-11} \times 20 \times 50}{ 9.8 \times 10^{-5}}}\) = 0.026 m
Ques. Define the Universal Law of Gravitation. (3 Marks)
Ans. According to the Universal Law of Gravitation, the force of attraction between two particles in the universe is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them.”
Mathematically,
F ∝ \(\frac{m_1m_2}{r^2}\)
Where
- F is the gravitational force
- m1 and m2 are the masses of the two objects
- r is the distance between the bodies
On removing the proportionality sign, we get
F = G \(\frac{m_1m_2}{r^2}\)
Where G is a constant of proportionality, known as Gravitational constant.
Ques. Write the dimensional formula of the Gravitational constant. (1 Mark)
Ans. The dimensional formula of the Gravitational constant is [ M-1 L3 T-2 ].
Ques. Two solid spheres of the same density and same radius r are placed in contact with each other. Show that the gravitational force between them is directly proportional to the fourth power of radius r. (5 Marks)
Ans. Since the density and radius of both the spheres are same, therefore their mass will be equal. Let the mass of each sphere be m.
Also, both the spheres are placed in contact with each other, therefore the distance between their center of mass will be 2r.
The gravitational force between the spheres will be
F = G \(\frac{m \times m}{(2r)^2}\) = G \(\frac{m^2}{4r^2}\)
But, the mass of each sphere, m = volume of the sphere (V) x density of sphere (ρ)
Volume of the sphere, V = 4/3 πr3
⇒ m = \(\frac{4}{3}\) π r3 ρ
⇒ F = G\(\frac{(\frac{4}{3} \pi r^3)^2}{4r^2}\) = \(\frac{4}{9}\) π2Gρ2r4
Since \(\frac{4}{9}\)π2Gρ2 = constant, therefore
F ∝ r4
Hence it is proved that the gravitational force between them is directly proportional to the fourth power of radius r.
Ques. A body weighs 36 kg on the surface of the earth. How much will it weigh on the surface of Mars whose mass is 1/9 and radius is 1/2 of that of the earth? (5 Marks)
Ans. We know that the weight (We) of the body on the surface of the earth is equal to the force of attraction (F) between the earth and the body. i.e.
We = mg = GMem/Re2
Where
- Me = mass of the earth
- Re = radius of the earth
- m = mass of the body
- g = acceleration due to gravity
- G = gravitational constant
Let Mm and Rm be the mass and radius of the Mars respectively.
Given
- Mass of the mars, Mm = 1/9 Me
- Radius of the mars = 1/2 Re
If Wm be the weight of the body of the surface of the mars, then
Wm = G \(\frac{M_mm}{R_m^2}\) = G\(\frac{(M_e/9)^m}{(1/2R_e)^2}\) = \(\frac{4}{9}\) G \(\frac{M_mm}{R_e^2}\)
⇒ Wm = (4/9) We
Given, the weight of the body of the surface on the earth, We = 36 kg
Then, the weight of the body on the surface of the mass, will be
Wm = (4/9) x 36 = 16 kg
Ques. A sphere of 10 kg is attracted by a second sphere of mass of 60 kg with a force equal to 0.02 gf. If G = 6.67 x 10-11 N m2 kg-2, calculate the distance between them. (3 Marks)
Ans. Given
- Mass of the first sphere, m1 = 10 kg
- Mass of the second sphere, m2 = 60 kg
- Gravitational force between the spheres, F = 0.02 gf = 19.6 x 10-5 N
The formula for gravitational force between two bodies is given by
F = G m1m2/r2
Where r is the distance between the two bodies.
Therefore,
r = \(\sqrt{G\frac{m_1m_2}{F}}\) = \(\sqrt{\frac{6.67 \times 10^{-11} \times 10 \times 60}{ 19.6 \times 10^{-5}}}\) = 0.014 m
Ques. A body weighs 54 kg on the surface of the earth. How much will it weigh on the surface of Mars whose mass is 1/9 and radius is 1/2 of that of the earth? (5 Marks)
Ans. We know that the weight (We) of the body on the surface of the earth is equal to the force of attraction (F) between the earth and the body. i.e.
We = mg = G Mem/Re2
Where
- Me = mass of the earth
- Re = radius of the earth
- m = mass of the body
- g = acceleration due to gravity
- G = gravitational constant
Let Mm and Rm be the mass and radius of the Mars respectively.
Given
- Mass of the mars, Mm = 1/9 Me
- Radius of the mars = 1/2 Re
If Wm be the weight of the body of the surface of the mars, then
Wm = G\(\frac{M_mm}{R_m^2}\) = G\(\frac{(M_e/9)^m}{(1/2R_e)^2}\) = \(\frac{4}{9}\) G \(\frac{M_mm}{R_e^2}\)
⇒ Wm = (4/9) We
Given, the weight of the body of the surface on the earth, We = 54 kg
Then, the weight of the body on the surface of the mass, will be
Wm = (4/9) x 54 = 24 kg
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