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We define the acceleration due to gravity as the constant acceleration produced on the body when it freely falls under the effect of gravity. It is denoted by \[\overrightarrow{g}\] because it has both direction and magnitude.

Let us say a body of mass m is placed at distance R from the Earth whose mass is M.

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We know that the earth tries to pull everything towards it by force. So, it pulls the body by force, mg.

Now, using the law of gravitation, we get,

\[\frac{GMm}{R^{2}} = mg\]

\[g = \frac{GM}{R^{2}}\]

This is the formula for the acceleration due to gravity.

Let us say, a person of mass m is standing at point B at a distance ‘R’ from the center of the Earth whose mass is M.

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According to the Gauss law of gravitation, the Earth is a symmetric solid sphere whose mass is concentrated at the center. If we wish to calculate the gravitational field strength \[\overrightarrow{E}\] at point B is \[\frac{GM}{R^{2}}\].

This \[\overrightarrow{E}\] is directed towards its mass, i.e, at the center. This value remains the same for a person standing outside and at the surface of the Earth.

Here, we can see that \[\overrightarrow{E} \propto \frac{1}{R^{2}}\]. We can show this by a graph.

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This means the value of \[\overrightarrow{E}\] is maximum at the surface and decreases with the increase in the distance from the earth.

We know that the earth’s gravitational field is \[\overrightarrow{g}\]. So, \[\overrightarrow{E} = \overrightarrow{g}\].

From the formula, \[g = \frac{GM}{R^{2}}\], we can see that \[\overrightarrow{E} \propto \frac{1}{R^{2}}\].

This means the value of g varies with the variation in R.

It is maximum at the surface of the earth because of the distance of a person from the center of the earth, which is equal to the radius of the earth Re.

Now, R becomes equal to Re.

So, the formula for g becomes: \[g = \frac{GM}{R_{e}^{2}}\].

∵ g is maximum at the surface and decreases with the increase in the distance from the earth.

∴ The graph will be the same for g, as we have drawn for \[\overrightarrow{E}\].

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We know that formula for the acceleration due to gravity on the surface of the earth is:

\[g = \frac{GM_{e}}{R_{e}^{2}}\].

Where,

\[G = \text{Universal gravitational constant; whose value is } 6.673 \times 10^{-11} Nm^{2}kg^{-1}\]

\[M_{e} = 6 \times 10^{24} kg\]

\[R_{e} = 6.4 \times 10^{6}m\]

On putting these values in the above formula, we get the value of g at the surface of the earth as 9.8 ms^{-2}. However, this value varies with height.

The dimensional formula for g is \[[M^{0}LT^{-2}]\].

Consider earth of mass M, radius Re with center at O.

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Let g = acceleration due to gravity at its surface, then \[g = \frac{GM}{R_{e}^{2}}\] …(1)

If gh is the acceleration due to gravity at a point h above the surface of the earth, then

\[g_{n} = \frac{GM}{(R + h)^{2}}\]…(2)

Now, (2) ÷ (1)

\[\frac{gh}{g} = \frac{R_{e}^{2}}{(R_{e} + h)^{2}} = (1 + \frac{h}{R_{e}})^{-2}\]

∵ h < < Re, therefore, h/R is a very small value. Now expanding the term (1 + h/R_{e})^{2} by Binomial theorem, we get,

We can see that \[g_{e} \propto \frac{1}{R_{e}}\], the value of g decreases with height.

From the formula \[g = \frac{GM_{e}}{R_{e}^{2}}\], we can see that as we go below the earth’s surface, the distance between us and the center of the earth decreases. Therefore, the value of g decreases. Now, let us understand how it decreases.

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Let us say, you are standing at point R and the earth is attracting you by force mg’. Here, we are using g’ in place of g because we have come inside the earth.

Using the law of gravitation

\[mg' = \frac{GM}{R^{2}}\]

If you observe, you are attracted by the inner surface, while M is the mass of the earth. So, we will use another method here:

If ρ is the uniform density of the inner surface, then

\[M = \frac{4}{3} \pi R^{3} \rho\]

\[\text{So, } g = \frac{G \frac{4}{3} \pi R^{3} \rho}{R^{2}}\] ..(1)

Let the mass of the inner surface be M’, and radius, R - d, then

\[g’ = \frac{GM’}{(R - d)^{2}}, \text{and M’ } = \frac{4}{3} \pi (R - d)^{3} \rho\]

\[g' = \frac{G\frac{4}{3} \pi {(R-d)}\rho ^{3}}{(R-d)^{2}}\]…(2)

Now, (3) ÷ (2)

\[\frac{g’}{g} = \frac{\frac{G\frac{4}{3} \pi {(R-d)}\rho ^{3}}{(R-d)^{2}}}{\frac{G \frac{4}{3} \pi R^{3} \rho}{R^{2}}}\]

On solving, we get,

∴ The value of g decreases with depth.

FAQ (Frequently Asked Questions)

Q1: Find the Height above the Earth’s Surface Where g Drops to 64% of g at the Surface.

Ans: Here, g_{h} = 64/100 g

Now, using the formula, g_{h} = g (1 + h/R_{e})^{2}

64/100 g = g/(1 + h/R_{e})^{2}

(1 + h/R_{e})= 10/8

h = 1/4 R_{e}

Q2: Why is the Value of g Maximum at Pole and Minimum at the Equator?

Ans: The expression for the value of g due to the rotation of the earth is:

g' = g - Rω^{2}Cos^{2}ф

Here, ф is the latitude.

At the equator, ф = 0°, Cos ф = Cos 0° = 1, then

g' = g - Rω^{2}

∴ g’ is minimum at the equator.

At the pole, ф = 90°, Cos ф = Cos 90° = 0, then

g' = g - Rω^{2}(0)^{2}

g’ = g, which is maximum

Hence, the value of g is minimum at the equator and maximum at the pole.

Q3: Where is the Gravity Strongest?

Ans: The gravity is strongest at the surface of the earth.

Q4: What does Microgravity Mean?

Ans: Microgravity is a condition in which astronauts feel weightless because of the absence of the force of gravity outside the earth.