Acceleration due to Gravity: Value of g, Escape Velocity

A-falling object is an object that is falling solely under the influence of gravity. Such an object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value is so important that it is given a special name. It is known as acceleration due to gravity.

Acceleration due to Gravity is the acceleration due to the force of gravitation of the earth. A parachutist jumping from an aeroplane fallsly for some time. He experiences an acceleration caused by the earth’s gravity. Let us see more about gravity and its relationship with acceleration in this article.

Gravity Definition

In the universe, all objects have the gravitational force and pull each other. This force due to the earth is called gravity. Just as the earth pulls us towards itself, we too pull it towards ourselves. However, due to the enormous difference in masses, the effect of our pull is negligible.

What is Acceleration?

We define acceleration as the rate of change of velocity. It means that acceleration is a change in velocity divided by time. When an object is in motion, it can change velocity due to force on it.

If the initial velocity was \(u,\) and after \(t\) seconds, it changed to velocity, \(v,\) then this means that- \({\text{acceleration}}\left( a \right) = \frac{{{\text{final}}\,{\text{velocity}}\left( v \right) – {\text{initial}}\,{\text{velocity}}\left( u \right)}}{{{\text{time}}\left( t \right)}}\) Unit of acceleration is metre per second square, \({\text{m}}/{{\text{s}}^2}.\)

What is Acceleration Due to Gravity in Simple Words?

Aly falling body experiences a force that of gravity. This force increases the velocity of the falling object at a constant rate. This constant rate of change of velocity of aly falling object is known as acceleration due to gravity.

Derivation of Acceleration Due to Gravity

The universal law of gravitation states that two bodies are attracted to each other by a force which is-

i. directly proportional to the product of their masses, and
ii. inversely proportional to the square of the distance \(d\) between their centres

Let us consider the mass of earth as \(M\) and that of an object is \(m.\)
Gravitational force on the object of mass \(m,\)

\(F \propto M \times m\)
\(F \propto \frac{1}{{{d^2}}}\)
Combining them, \(F \propto \frac{{Mm}}{{{d^2}}}\)
\(F = \frac{{GMm}}{{{d^2}}}\)

\(G\) is called the universal gravitational constant and is equal to \(6.67 \times {10^{ – 11}}\,{\text{N}}{{\text{m}}^2}/{\text{k}}{{\text{g}}^2}.\)

Force is defined as the product of mass and acceleration.

\({\text{Forc}}e = {\text{mass}} \times {\text{acceleration}}\)
\(F = ma\)
For gravity it is- \(F = mg\)
Therefore, \(F = mg = \frac{{GMm}}{{{d^2}}}\)
\(g = \frac{{GM}}{{{d^2}}}\)

For an object falling towards the earth, the distance between it and the surface is negligible compared to the radius \(r\) of the earth. So, we can equate \(d\) to \(r.\)
\(g = \frac{{GM}}{{{r^2}}}\)

Therefore, \(g\) depends on the-
i. mass of the earth, \(M\)
ii. the radius of the earth, \(r\)
The mass of the object does not affect \(g.\) All objects fall at the same acceleration simultaneously from the same height, irrespective of their weights. This was demonstrated by the Italian astronomer and scientist Galileo Galilei from the Leaning Tower of Pisa.

If a leaf or piece of paper falls slowly, it is because of air resistance to such light objects. Air resistance makes a parachute slow down the fall. A feather and a block of iron fall to the ground simultaneously in a vacuum.

Value of Acceleration Due to Gravity

Universal gravitational constant, \(G = 6.67 \times {10^{ – 11}}\,{\text{N}}{{\text{m}}^2}\,{\text{k}}{{\text{g}}^{ – 2}}\)
Mass of the earth, \(M = 6 \times {10^{24}}\,{\text{kg}}\)
The average radius of the earth, \(r = 6.4 \times {10^6}\,m\)
\(\therefore \,g = \frac{{GM}}{{{r^2}}} = \frac{{6.67 \times {{10}^{ – 11}} \times 6 \times {{10}^{24}}}}{{{{\left( {6.4 \times {{10}^6}} \right)}^2}}} = 9.8\;{\rm{m}}/{{\rm{s}}^{ – 2}}\)
Aly falling body increases its velocity by \(9.8\,{\text{m}}/{\text{s}}\) every second as it nears the surface.

Acceleration Due to Gravity in Terms of Density

The density of an object is the ratio of its mass to its volume.

Let us assume the earth to be a perfect sphere. So, its volume is:

\(V = \frac{4}{3}\pi {r^3}\)
Density \(\rho = \frac{M}{V}\)
\(g = \frac{{GM}}{{{r^2}}} = \frac{G}{{{r^2}}}\rho V = \frac{{G\rho }}{{{r^2}}} \times \frac{4}{3}\pi {r^3}\)
\(g = \frac{{4\pi }}{3}G\rho r\)

Acceleration due to gravity is directly proportional to the density of the earth.

Factors Affecting Acceleration Due to Gravity

Three factors affect acceleration due to gravity. They are:
i. The shape of the earth.
ii. Altitude or height.
iii. Depth.

Variation in \(g\) Due to the Shape of the Earth

The earth is not a perfect sphere but is bulged at the equator and slightly flattened at the poles. So, the value of \(g\) is somewhat more at the poles compared to the equator.

Variation in \(g\) Due to Altitude

Let \({g_h}\) be the acceleration due to gravity at a height \(h\) above the surface of the earth.
\(g = \frac{{gM}}{{{r^2}}}\)
\({g_h} = \frac{{GM}}{{{{\left({r + h} \right)}^2}}}\)
Dividing \({g_h}\) by \(g\)
\(\frac{{{g_h}}}{g} = \frac{{{r^2}}}{{{{\left({r + h} \right)}^2}}}\)
\({g_h} = \frac{g}{{{{\left({1 + \frac{h}{r}} \right)}^2}}}\)

This equation shows that as \(h\) increases, the value of \({g_h}\) decreases. At a considerable height, \(h > > r,\) and acceleration due to gravity becomes negligibly small. The earth’s gravity does not influence objects there.

Variation in \(g\) Due to Depth

Let \({g_d}\) be the acceleration due to gravity at a depth, \(d\) below the surface of the earth. Since it is below the surface, we must consider the acceleration in terms of density.
\(g = \frac{{4\pi }}{3}G\rho r\)
\({g_d} = \frac{{4\pi }}{3}G\rho \left({r – d} \right)\)
Dividing \({g_d}\) by \(g\) \({g_d} = \frac{{\left({r – d} \right)}}{r}\)
\({g_d} = g\left({1 – \frac{d}{r}} \right)\)
At surface, \(d = 0,{g_d} = g\)
At the centre of the earth, \(d = r.\) So, \({g_d} = 0.\) Acceleration due to gravity at the centre of the earth is zero.

Difference Between Mass and Weight

Mass and weight sound similar but they represent different physical quantity. Following are the differences between them.

Since weight is a force due to gravity, it has the formula,
Weight,\(W = mg\)
Where \(g\) is the acceleration due to gravity at that place.

Comparison of Weight on Earth and on Moon

The mass of the earth is approximately \(100\) times that of the moon. The radius of the earth is about \(4\) times that of the moon.
Acceleration due to gravity for the earth:
\(g = \frac{{GM}}{{{r^2}}}\)
Acceleration due to gravity for the moon:
\({g_o} = \frac{{G{M_o}}}{{{r_o}^2}}\)
Given,
\(M = 100\,{M_o}\)
Dividing,
\(r = 4{r_o}\)
\(\frac{g}{{{g_o}}} = \frac{M}{{{M_o}}} \times {\left({\frac{{{r_o}}}{2}} \right)^2} = 100 \times {\left({\frac{1}{4}} \right)^2} = 6.25\)
Since \(m\) remains the same both on the Earth and the Moon,
\(\frac{W}{{{W_o}}} = 6.25\)
Weight on an object is approximately \(6\) times that on the moon, or weight on the moon is \({\left( {\frac{1}{6}} \right)^{th}}\) of that on earth.

Acceleration Due to Gravity – Sample Problems

Q.1. The mass of the earth is \(10\) times that of another planet. Radius of the earth is \(2\) times its radius. Compare the weight of an object on the planet with that on the earth.
Ans: Given,
\(M = 10{M_p}\)
\(r = 2{r_p}\)
Dividing,
\(\frac{g}{{{g_p}}} = \frac{M}{{{M_p}}} \times {\left({\frac{{{r_p}}}{r}} \right)^2} = 10 \times {\left({\frac{1}{2}} \right)^2} = 2.25\)
Since weight \(W = mg,\) and m remains the same everywhere,
\(\frac{W}{{{W_p}}} = 2.25\)
Weight on the planet is approximately half of that on earth.

Q.2. Two balls of equal mass \(2\) kilograms are separated by \(1\) metre from their centres. What is the force of attraction between them?
Ans: \(F = \frac{{GMm}}{{{d^2}}}\)
Universal gravitational constant, \(G = 6.67 \times {10^{ – 11}}\,{\text{N}}{{\text{m}}^2}\,{\text{k}}{{\text{g}}^{ – 2}}\)
\(F = \frac{{6.67 \times {{10}^{ – 11}} \times 2 \times 2}}{{{1^2}}}\)
\(F = 2.668 \times {10^{ – 10}}\,{\text{N}}\)

Summary

a. Acceleration due to gravity \(\left( g \right)\) near the surface of earth is \(9.8\,{\text{m}}/{{\text{s}}^2}.\) All thefalling objects fall towards earth with this acceleration.
b. \(g\) decreases with an increase in height above the surface.
c. \(g\) decreases with an increase in depth below the surface.
d. Mass is the measure of amount of substance present in an object, whereas weight is the force due to gravity on it. Mass is a scalar, and weight is a vector.

Acceleration Due to Gravity FAQs

The most frequently asked questions about acceleration due to gravity are answered here:

Q.1. Where is acceleration due to gravity is maximum?
Ans: Acceleration due to gravity is inversely proportional to the square of the distance between the centre and the surface. Poles are closer to the centre than the equator. So, acceleration due to gravity is more at the poles.

Q.2. Why do we say that unit of weight is Newton?
Ans: Weight is the force due to the mass acting downwards. So, its unit is Newton.

Q.3. Why cannot aeroplanes be used to go into outer space?
Ans: A minimum velocity is required to escape from earth’s gravitational force, called escape velocity. Aeroplanes cannot travel at this velocity.

Q.4. Is the gravity 9.8ms–29.8ms–2?
Ans: The acceleration due to gravity on the surface of the earth is 9.8ms–2.9.8ms–2. Gravity is the force exerted by the Earth on objects present on and around it.

Q.5. Why is acceleration due to gravity is 9.8ms–29.8ms–2?
Ans: Acceleration due to gravity g=GMR2.g=GMR2. Substituting G=6.67×10–11Nm2kg–2,R=6.4×106mG=6.67×10–11Nm2kg–2,R=6.4×106m and M=6×1024kg,M=6×1024kg, we get g≈9.8ms–2