Acceleration due to Gravity
By The acceleration due to gravity at a height $h$ above the earth surface is given by \begin{align} g=\frac{GM}{ (R+h)^2} \end{align} where $G$ is universal gravitational constant, $M$ is mass of the earth and $R$ is earth's readius (assuming spherical). The acceleration due to gravity on the earth's surface (at equator) is approximately $g_e=9.8$ m/s2.
The variation of acceleration due to gravity $g$ with distance $r$ from centre of the earth is given by \begin{align} g=\left\{\begin{array}{cc} \frac{GMr}{R^3}, &\text{(for $r < R$)};\\ \frac{GM}{r^2}, &\text{(for $r > R$)}. \end{array}\right. \end{align}
The acceleration due to gravity at a depth $h\,(\ll R)$ below the earth surface can be approximated as \begin{align} g_\text{inside}\approx g\left(1-\frac{h}{R}\right) \end{align}
The acceleration due to gravity at a height $h\,(\ll R)$ can be approximated as \begin{align} g_\text{outside}\approx g\left(1-\frac{2h}{R}\right) \end{align}
The acceleration due to gravity at the pole is greater than that at the equator due non-spherical shape of the earth.
Effect of earth rotation on apparent weight: \begin{align} mg^\prime_\theta=mg-m\omega^2 R\cos^2\theta \end{align}
Problems from IIT JEE
Problem (JEE Mains 2009): The height at which the acceleration due to gravity becomes $g/9$ (where $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is
- $2R$
- $R/\sqrt{2}$
- $R/2$
- $\sqrt{2}R$
Solution: The acceleration due to gravity at a height $h$ above the earth's surface is given by, \begin{align} g^\prime &=\frac{GM}{(R+h)^2} \\ &=\frac{GM}{R^2}\cdot\frac{R^2}{(R+h)^2} \\ &=\frac{gR^2}{(R+h)^2}.\nonumber \end{align} Substitute $g^\prime=g/9$ and solve to get $h=2R$.
Problem (JEE Mains 2009): The ratio of the weights of a body on the earth's surface to that on the surface of a planet is $9:4$. The mass of the planet is $\frac{1}{9}$th of that of the earth. If $R$ is the radius of the earth, what is the radius of the planet? (Take the planets to have the same mass density.)
- $R/3$
- $R/4$
- $R/9$
- $R/2$
Solution: Let $M$ be the mass of the earth. The mass of the planet is $M_p=M/9$. Let $R_p$ be the radius of the planet. The ratio of the weights of a body (of mass $m$) on the earth's surface to that on the planet's surface is, \begin{align} \frac{W}{W_p}&=\frac{mg}{mg_p} \\ &=\frac{g}{g_p}\\ &=\frac{GM/R^2}{GM_p/R_p^2}\\ &=\frac{GM/R^2}{G(M/9)/R_p^2}\nonumber\\ &=\frac{9R_p^2}{R^2}\\ &=\frac{9}{4}.\nonumber \end{align} Solve to get $R_p=R/2$.
This problem from JEE Mains 2019 also states: "Take the planets to have the same mass density". Can you see that this overstatement is wrong?
Problem (IIT JEE 1984): The numerical value of the angular velocity of rotation of the earth should be__________rad/s in order to make the effective acceleration due to gravity at the equator equal to zero.
Solution: The effective acceleration due to gravity at the equator is given by \begin{align} %\label{ymb:eqn:1} g^\prime=g_0-\omega^2 R,\nonumber \end{align} where $g_0=9.8$ m/s2 is a constant, $\omega$ is the angular velocity of the earth and $R=6400\,\mathrm{km}=6.4\times{10}^{6}$ m is the radius of the earth. Substitute the values to get $\omega$ at which $g^\prime$ becomes zero i.e., \begin{align} \omega&=\sqrt{g_0/R}\\ &=1.24\times{10}^{-3} \;\mathrm{rad/s}. \end{align}