Relation Between G and g

The relation between g (acceleration due to gravity) and G (universal gravitational constant) forms a fundamental concept in gravitational physics. G is a constant that defines the strength of the gravitational force between two masses in the universe, while g represents the acceleration experienced by an object due to Earth's gravitational pull. The relation between the two can be expressed mathematically using Newton's law of gravitation or the Universal Law of Gravitation, which helps in understanding how the mass and radius of a celestial body, like Earth, influence the gravitational acceleration experienced at its surface.

Let's read this article to learn about the difference between capital g and small g, and the value of G on different surfaces. This is one of the very important concept for Class 12 Board exams and competitive exams like JEE, NEET and other state engineering exams.

Acceleration Due to Gravity (g) on Earth

An object which falls under the sole influence of our gravity is called a free-falling object or acceleration due to gravity.
A free-falling object usually has an acceleration of about $9.8 \mathrm{~m} \mathrm{~s}^{-2}$, downward towards the earth. We usually denote it with the symbol g.

• Acceleration due to gravity g on earth formula = g = $9.8 \mathrm{~m} \mathrm{~s}^{-2}$
• Acceleration due to gravity dimensional formula = $M^0 L^1 T^{-2}$
• SI Unit of g = m/s

Also read -

• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for All Subjects

Universal Gravitational Constant (G)

The force of attraction which is given between two unit masses that are separated by some unit distance is called the universal gravitational constant. This universal gravitational constant is often denoted by the symbol G (capital G) and is measured in $\mathrm{Nm}^2 / \mathrm{kg}^2$.

• Value of gravitational constant G = $6.67 \times 10-11 \mathrm{Nm}^2 / \mathrm{Kg}^2$

= $6.67 \times 10^{-8}$ dyne $\mathrm{cm}^2 \mathrm{~g}^{-2}$

= $6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{Kg}^{-2}$

• Dimensional formula for G = $M^{-1} L^3 T^{-2}$

Difference Between g and G

Relation Between G and g

After knowing about the difference between capital g and small g, move to the main part of the article which is the relation between capital g (G) and small g (g).

Let's suppose the mass of the earth is M and its radius is R, then the force of attraction on the given body is equal to the surface of the earth is given by

$F=\frac{G M m}{R^2}$

Here,

• g →acceleration due to the gravity measured in m/s 2 .
• G →universal gravitational constant measured in Nm 2 /kg 2 .
• R →radius of the massive body measured in km.
• M →mass of the massive body measured in Kg

Now, from the second law of motion

$g=\frac{F}{m}=\frac{G M}{R^2}$ ---- (1)

We can see the above equation is free from the mass(m), which means we can say acceleration due to gravity does not depend on the mass. Hence it is equal for all.

Now, to get the value of g we need to put the value of the known parameter in equation (1).

$G= 6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}, M$ as $6 \times 10^{24} \mathrm{~kg}$ and $R$ as $6.4 \times 10^6 \mathrm{~m}$,

After putting and solving the equation (1), we will get the value of g= 9.8 m s -2

Also Read:

• NCERT Solutions for Class 11 Physics Chapter 8 Gravitation
• NCERT Exemplar Class 11 Physics Solutions Chapter 8 Gravitation
• NCERT notes Class 11 Physics Chapter 8 Gravitation

Variation in the Value of Acceleration Due to Gravity (g)

In the above section we have seen the relation between g and G, Now let's move to a little depth of how the g value is changing in different cases.

It may vary due to the following factors

• Height above the surface of the earth
• Depth below the surface of the earth
• Shape of the earth

Let's read one by one starting from height above the surface of the earth.

(i) Height above the surface of the earth

Suppose a body of mass m is lying on the surface of the earth and its radius is R. Then its acceleration due to gravity at the earth's surface,

$g=\frac{G M}{R^2}$

Height h above the surface, the force of gravity,

$F=\frac{G M m}{(R+h)^2}$

Therefore, the acceleration due to gravity at height h above the surface

$g^{\prime}=\frac{F}{m}=\frac{G M}{(R+h)^2}$

Or,

$
g' = \frac{G M}{R^2 \left(1 + \frac{h}{R}\right)^2} \Rightarrow g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \quad \left(\because \frac{G M}{R^2} = g\right)
$

Thus,

$
g' < g
$

(ii) Depth below the surface of the earth

Now for this situation, let's suppose an object of mass m is kept at a depth of d below the earth's surface. Its distance from the centre of the Earth is (R − d). This mass is located on the outer surface of the Inner solid sphere. Gravitational pull on a The mass within a spherical shell is always zero. Consequently, the Objects experience gravitational attraction mainly because Inner solid sphere

Mass of this sphere $M^{\prime}=\left\{\frac{M}{(4 / 3) \pi R^3}\right\} \frac{4}{3} \pi(R-d)^3$

Here, M represents the mass of the entire sphere of radius R.

Therefore,

$M^{\prime}=\frac{(R-d)^3}{R^3} M$

Now, the force of gravity on object at depth d

$
F = \frac{G M' m}{(R - d)^2} = \frac{G M m (R - d)}{R^3} \quad \text{and} \quad g' = \frac{F}{m} = \frac{G M}{R^3}(R - d)
$

Since,

$
g = \frac{G M}{R^2} \quad \text{-----(1)}
$

After substituting the value in equation (1),

$g^{\prime}=g\left(1-\frac{d}{R}\right)$

(iii) Shape of the earth

The earth is not a perfect sphere. It is somewhat flat at the two poles. The equatorial radius (R eq ) is approximately 21 km more than the polar radius ( R p ).

$
g_p = \frac{G M}{R_p^2}
$

Acceleration due to gravity at the equator,

$
\begin{aligned}
g_{\mathrm{eq}} & = \frac{G M}{R_{\mathrm{eq}}^2} \\
\because R_p & < R_{\mathrm{eq}} \\
\therefore g_p & > g_{\mathrm{eq}}
\end{aligned}
$

In short, by going through this article which starts with the difference between capital g and small g, and covers the relation between G and g along with the last value of g in different surfaces you have a solid understanding of this concept. Solve numerical types of questions based on this concept to boost your confidence before the final board exam or any competitive exams.

Also check-

• NCERT Exemplar Class 11th Physics Solutions
• NCERT Exemplar Class 12th Physics Solutions
• NCERT Exemplar Solutions for All Subjects

NCERT Physics Notes:

• NCERT Notes Class 11th Physics
• NCERT Notes Class 12th Physics
• NCERT Notes For All Subjects