JEE Main 2023 — Gravitation Question with Solution

The acceleration due to gravity ($g$) at the surface of Earth is approximately $9.81{\text{m/s}}^2$. This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth. This force decreases with the square of the distance from the center of the Earth, due to the inverse square law.

However, if we are at a height ($h$) much less than the radius of the Earth ($R$), we can use a linear approximation to calculate the new acceleration due to gravity ($g^{\prime }$) at this height. This is because for small heights compared to the radius of Earth, the decrease in $g$ can be approximated to be linear. This is represented by the following formula:

$$g^{\prime }=g\left(1-\frac{2h}{R}\right)$$

Here's how this formula is derived:

The force of gravity ($F$) is given by the universal law of gravitation:

$$F=G\frac{m_1{m}_2}{d^2}$$

where:

• $G$ is the gravitational constant,
• $m_1$ and $m_2$ are the masses of the two objects (in this case, the Earth and the object we're considering),
• $d$ is the distance between the centers of the two objects.

If we consider an object of mass $m$ at the surface of Earth, the force it experiences due to gravity is:

$$F=G\frac{mM}{R^2}=mg$$

where:

• $M$ is the mass of Earth,
• $R$ is the radius of Earth,
• $g=\frac{GM}{R^2}$ is the acceleration due to gravity at the Earth's surface.

Now, if the object is at a height $h$ above the Earth's surface, the force it experiences is:

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

We can write $(R+h{)}^2$ as $R^2(1+\frac{h}{R}{)}^2$, and because $h<<R$, we can use the binomial approximation $(1+x{)}^2\approx 1+2x$ for small $x$ to get:

$$F^{\prime }\approx G\frac{mM}{R^2}\left(1-\frac{2h}{R}\right)=mg\left(1-\frac{2h}{R}\right)$$

Equating the forces $F=mg$ and $F^{\prime }=m{g}^{\prime }$ gives us:

$$g^{\prime }=g\left(1-\frac{2h}{R}\right)$$