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Gravitational constant or g is the proportionality constant used in Newton’s Law of Gravitation. Acceleration due to gravity, represented by g, is the acceleration experienced by a freely falling object as a result of the gravitational force. The SI unit used with the value of g is m/s2. The gravitational force between two objects is equal to the product of their masses and inversely proportional to the square of the distance between their centres. The value of g is dependent on the mass of the massive body and its radius. The value of g varies from one body to another.
Key Terms: Gravitational Constant, Proportionality, Newton’s Law of Gravitation, Acceleration due to Gravity, Mass, Distance, Gravitational Force
Universal Gravitational Constant
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As per Newton's law, any two objects having mass m1 and m2 (in kilograms), with their centres separated by a distance r (in meter), will have a gravitational force F (in Newton) to exist between them, This force is denoted by:
Fgrav = m*g
F = Gm1m2/r 2
- The force of attraction between any two unit masses separated by a unit distance is called the Universal Gravitational Constant denoted by G measured in Nm2/kg2.
- It is an empirical physical constant used in gravitational physics. It is also called Newton's Constant.
- Everywhere in the cosmos, the gravitational constant has the same value.
- G is not the same as g as the latter represents acceleration due to gravity.
Acceleration due to gravity on Earth
The acceleration due to gravity on Earth or the value of g on Earth is 9.8 m/s2. Therefore, the velocity of an object undergoing free fall increases by 9.8 each second. This acceleration is a result of the Earth’s gravity.
The video below explains this:
Relation Between G And g Detailed Video Explanation:
Read More: Kinetic Energy
Gravitational Constant - Derivation
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The value of the gravitational constant G in the Universal law of gravitation can be established empirically.
Let us consider two bodies of mass m1 and m2. Let r be the distance between the center of these bodies and is inversely proportional to the square of the distance between them. The force between the two bodies is given by,
F ∝ m1.m2 / r2
or,
F = G.m1.m2 / r2
Here, G is the constant of proportionality called the universal gravitational constant.
Let, m1 = m2 =1 and r =1
Then the equation becomes,
→ F = G*1*1/1
→ F = G
- Unit of F → Newton(N)
- Unit of mass → Kg
- Unit of R → m
- Unit of G → Nm²Kg²
Thus, we can conclude that the Universal gravitational constant is equal to the force of attraction acting between two unit mass bodies when their centers are placed a unit distance apart.
The currently accepted value of G is
G = 6.67×10-11 N m2/kg2
Read More: Potential Energy
Value of G
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In physics, we use two g, one which is small letter g is the acceleration due to gravity and the capital letter G is the universal gravitational constant.
- The value of the gravitational constant remains unchanged on the moon, Mars, or anywhere else in the universe, making it an invariant entity.
- According to the common Big Bang hypothesis and some astronomers, as the universe expands, the value of G will eventually decrease.
| Important Links Related to Gravitational Constant | ||
|---|---|---|
| Value of g on Moon | Escape Velocity | Gravitation Vs Gravity |
| Universal Gravitational Formula | Gravitational Potential Energy | G and g Relation |
| Escape Velocity vs Orbital Velocity | Weightlessness | Gravitational Force and Law of Gravitation |
Measuring Gravitational Constant
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Gravity is one of nature's four basic forces. Some of the methods to measure gravitational constant are –
- In the first technique, the researchers created a metal-coated silica plate that was suspended in the air by a wire. The gravitational attraction is provided by the two steel balls. The force of gravity was calculated by determining how much the wire was twisted.
- The second technique was identical to the first, but the plate was suspended from a revolving turntable, which held the wire in place. The gravitational force was measured using this approach by observing the spin of the turntable. By using seismic characteristics in both techniques, the researchers were able to avoid influence from surrounding objects and disturbances.
Applications of Gravitational Constant
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Some of the applications of value of G include –
- The gravitational force between two planets can be determined accurately.
- It can be used to determine the gravitational force for objects near the earth such as satellites.
- Solar and lunar eclipses can also be predicted by the use of this gravitational constant.
- The value of G is helpful in calculating the trajectory of astronomical bodies and their motion.
Acceleration due to Gravity
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The net acceleration that is transmitted to objects owing to the combined action of gravitation (from mass distribution within Earth) and the centrifugal force (from Earth's rotation) is the acceleration due to gravity. It is indicated as g on Earth.
- The acceleration is measured in meters per second squared (m/s2) as per the SI unit or equally in Newtons per kilogram (N/kg or N.Kg-1).
- The gravitational acceleration near Earth's surface is approximately 9.81 m/s2.
- The speed of an object free-falling will increase by 9.81 meters per second every time.
- The acceleration due to gravity is denoted by ‘g’.
Differences between G and g
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The differences between Universal Gravitational Constant, G, and acceleration due to gravity, represented by g are as follows –
G vs g Differences
| Universal Gravitational Constant 'G’ | Acceleration Due to Gravity 'g' |
|---|---|
| G is a gravitational constant. | g is the acceleration due to gravity. |
| The universal gravitational constant describes the gravitational force between two bodies of unit mass separated by a unit distance. | The acceleration produced in a free-falling body owing to gravitational attraction is known as gravitational acceleration. |
| The value of G is constant everywhere on Earth and across the cosmos. | The value of 'g' varies depending on where you are on the planet. |
| Changes in height and depth from the earth's surface have no effect on the value of G. | The value of g decreases as we move deeper inside the Earth or higher from its surface. |
| The value of G is not zero in the center of the Earth or anyplace else. | The value of g at the earth's core is zero. |
| The value for G is 6.6734 × 10-11 N m2/kg2 throughout the universe. | The value is 9.8m/s2 on the surface of the Earth. |
| Its SI unit is N m2/kg2 | The SI unit is m/s2 |
Things to Remember
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- The Gravitational Constant is an empirical physical constant used in Newton's Law of Universal Constant to calculate gravitational effects.
- The value of g or the value of gravitational constant at any point in this universe. G = 6.67408×10-11Nm2/kg2.
- The dimensional formula for gravitational constant is [L]3[M]-1[T]-2
- The SI unit of gravitational constant is Nm2/kg2
Also Check Out:
| Important Topics From Chapter 8: Gravitation | ||
|---|---|---|
| Force of Attraction Formula | Neptune | Motion of Celestial Bodies in Space |
| Types of Forces | Mean Free Path | Average Acceleration Formula |
| Tangential Acceleration Formula | Inelastic collision | Acceleration Unit |
Previous Year Questions on Gravitational Constant
- What should be the radius for earth to become a black hole? [NEET 2014]
- Ratio of time periods of planets. [NEET 1994]
- Calculate the ratio of escape velocitites if radius of planet is doubled. [NEET 2013]
- What is the minimum speed of projection. [NEET 2013]
- Sort the kinetic energies of planet at different positions. [NEET 2018]
- Minimum velocity value for particle not to return. [NEET 2011]
- Velocity of the particle for the height to equal the radius. [NEET 2001]
- Calculate the radius of the geostationary satellite. [NEET 1992]
- Escape velocity when the mass and radius of earth is changed. [NEET 1997]
- Calculate the disctance of two planets from the sun. [NEET 1997]
- Escape velocity of planet with half the radius of the earth. [NEET 2000]
- Gravitational Force is required for? [NEET 2000]
- Gravitational force exerted on a body on the surface of the earth. [NEET 2000]
- How much will a body weigh at the centre of the earth. [NEET 2019]
- Magnitude of gravitational potential. [NEET 2011]
- Find the ratio of linear velocity. [NEET 2011]
- Find the acceleration of the satellite. [NEET 1994]
- Change in potential energy of an object at a height from the earth’s surface.[NEET 2019]
- Calculate the orbital speed. [NEET 1994]
- Choose the correct on gravitational constant. [NEET 2018]
Sample Questions
Ques. Let us assume that our galaxy consists of 2.5 x 1011stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 100000 ly. (3 marks)
Ans. Here, r = 50000 ly = 50000 x 9.46 x 1015 m = 4.73 x 1020 m
M = 2.5 x 1011 solar mass = 2.5 x 1011 x (2 x 1030) kg = 5.0 x 1041kg
We know that M = \(4 \pi^2r^3 \over GT^2\)
or T = \(({4 \pi^2r^3 \over GM})^{1/2}\)
M = \(4 \pi^2r^3 \over GT^2\)
or, T = \(({4 \pi^2r^3 \over GM})^{1/2}\) = \([\frac{4 * (22/7)^2 * (4.73 * 10 ^{20})^3}{(6.67 * 10^{-11} ) * (5.0 * 10^{41})}]^{1/2}\)
= 1.12 x 1016 s.
Ques.Does the escape speed of a body from the Earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? (3 marks)
Ans. The escape speed ves = \(\sqrt{\frac{2GM}{R}}\) = \(\sqrt{2gR}\) . Hence,
(a) The escape speed of a body from the Earth does not depend on the mass of the body.
(b) The escape speed does not depend on the location from where a body is projected.
(c) The escape speed does not depend on the direction of projection of a body.
(d) The escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.
Ques. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 x 1030 kg, mass of the earth = 6 x 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 1011 m).(3 marks)
Ans. Mass of Sun, M = 2 x 1030 kg; Mass of Earth, m = 6 x 1024 kg Distance between Sim and Earth, r = 1.5 x 1011 m
Let at the point P, the gravitational force on the rocket due to Earth
= gravitational force on the rocket due to Sun
Let, x = distance of the point P from the Earth
Then, \(\frac{Gm}{x^2} = \frac{GM}{(r-x)^2}\)
⇒ \(\frac{(r-x)^2}{x^2} = \frac{M}{m} = \frac{2*10^{30}}{6*10^{24}} = \frac{10^6}{3}\)
or, \(\frac{r-x}{x} = \frac{10^3}{\sqrt{3}}\)⇒ \(\frac{r}{x} \)= \( \frac{10^3}{\sqrt{3}}\) + 1 ≈ \( \frac{10^3}{\sqrt{3}}\)
or, x = \(\frac{\sqrt{3r}}{10^3} = \frac{1.732 1.5 * 10^{11}}{10^3} \)= 2.6 x 108 m.
Ques. How will you ‘weigh the sun’, that is, estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 108 km.(3 marks)
Ans. The mean orbital radius of the Earth around the Sun
R = 1.5 x 108 km = 1.5 x 1011 m
Time period, T = 365.25 x 24 x 60 x 60 s
Let the mass of the Sun be M and that of Earth be m.
According to law of gravitation F = G\(\frac{Mm}{R^2}\) …(i)
Centripetal force, F = \(\frac{mv^2}{R} \)= m.R.ω2…(ii)
From eqn.(i) and (ii), we have
\(\frac{GMm}{R^2}\) = m.R.ω2
= \(\frac{mR.4\pi^2}{T^2}\) [\(\because \) ω = \(\frac{2 \pi}{T}\)]
∴ M = \(\frac{4\pi^2R^3}{G.T^2}\)
= \(\frac{4*(3.14)^2 * (1.5 * 10^{11})^3}{6.67 * 10^{-11} * (365.25 * 24 * 60 * 60)^2}\)
= 2.009 x 1030 kg = 2.0 x 1030 kg.
Ques. A rocket is fired vertically with a speed of 5 km s-¹ from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 x 1024 kg; mean radius of the earth = 6.4 x 108 m; G = 6.67 x 10-11 N m2 kg-2(3 marks)
Ans. Initial kinetic energy of rocket = 1/2 mv2 = 1/2 x m x (5000)2 = 1.25 x 107 mJ
At distance r from centre of earth, kinetic energy becomes zero
.•. Change in kinetic energy = 1.25 x 107 – 0 = 1.25 x 107 m J
This energy changes into potential energy.
Initial potential energy at the surface of earth = GMem/’r
= \(\frac{-(6.67*10^{-11}) * (6 * 10^{24})m}{6.4 * 10^6}\) = – 6.25m x 107 J
Final Potential energy at distance, r = – GMem/r
= \(\frac{-(6.67*10^{-11}) * (6 * 10^{24})m}{r}\) = – 4 x 1014 \(\frac{m}{r}\)J
∴ Change in potential energy = 6.25 x 107 m – 4 x 1014 \(\frac{m}{r}\)
Using law of conservation of energy,
6.25 x 107 m – \(\frac{4*10^{14}m}{r} \) = 1.25 x 107 m
i.e, r = \(\frac{ 4 * 10^{14}}{5*10^7} m \)= 8x 1016 m.
Ques. The escape speed of a projectile on the Earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.(3 marks)
Ans. Let ves be the escape speed from surface of earth having a value ves = 11.2 kg s -1 ]
= 11.2 x 103 m.s-1. By definition
\(\frac{1}{2} mv_e^2 = \frac{GMm}{R^2}\)
When a body is projected with a speed vi = 3ves = 3 x 11.2 x 103 m/s, then it will have a final speed vf such that,
\(\frac{1}{2} mv_f^2 = \frac{1}{2} mv_i^2 - \frac{GMm}{R^2} = \frac{1}{2} mv_i^2 - \frac{1}{2} mv_e^2\)
⇒ vf = \(\sqrt{v_i^2 - v_e^2}\)
= \(\sqrt{(3*11.2*10^3) - (11.2*10^3)^2}\)
= 11.2 x 103 x \(\sqrt{8}\)
= 31.7 x 103 ms-1 or 31.7 km s-1.
Ques. Two heavy spheres each of mass 100 kg and radius 0.10 are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?(3 marks)
Ans. Here value of G = 6.67 x 10-11 Nm2kg-2 , M = 100 kg , R = 0.1 m, distance between the two spheres, d = 1.0 m
Suppose that the distance of either sphere from the mid-point of the line joining their center is r. Then r=d/2=0.5 m. The gravitational field at the mid-point due to two spheres will be equal and opposite.
Hence, the resultant gravitational field at the mis point =0
The gravitational potential at the mid point = ( – \(\frac{GM}{r} \)) x 2
= – \(\frac{6.67 * 10^{-11} * 100 * 2}{0.5} \)= – 2.668 x 10-8 J kg-1.
The object placed at that point would be in stable equilibrium.
Ques. A geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 x 1024 kg, radius = 6400 km.(3 marks)
Ans. Distance of satellite from the centre of earth = R = r + x
= 6400 + 36000 = 42400 km = 4.24 x 107 m
Using Potential, V = – \(\frac{GM}{R} \), we get
V = – \(\frac{(6.67*10^{-11}) * (6*10^{24})}{(4.24*10^7)} \) = – 9.44 x 106 J kg-1
Ques. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 x 1030 kg).(3 marks)
Ans. Acceleration due to gravity of the star, g = GM/R2 …………(i)
Here M is the mass and R is the radius of the star.
The outward centrifugal force acting on a body of mass m at the equator of the star =mv2/R =mR w2——-(ii)
From equation (i), the acceleration due to the gravity of the star
= \(\frac{6.67 * 10^{-11} * 2.5 * 2 * 10^{30}}{(12*10^3)^2}\) = 2.316 x 1012 m/s2
∴ Inward force due to gravity on a body of mass m
= m x 2.316 x 1012 N
From equation (ii), the outward centrifugal force = mRω2
= m x (12 x 103) x \((\frac{2\pi * 1.5}{-1})^2\)
= m x 1.06 x 106 N
Since the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.
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