The acceleration due to earth's gravity on a point particle at a height $h$ above the surface of the...

The gravitational acceleration experienced by a point particle at a height $ h $ above the Earth's surface, denoted as $ g_o(h) $, and at a depth $ d $ below the Earth's surface, denoted as $ g_i(d) $, are dependent on the Earth's radius $ R $ and its uniform mass density. The Earth's gravitational acceleration inside and outside its surface can be described mathematically.The gravitational acceleration inside the Earth at depth $ d $ is given by:$ g_i(d) = \frac{g}{R}(R-d) $The gravitational acceleration at a point outside the Earth at height $ h $ (often confused with depth $ d $ here trivialized) is calculated using:$ g_o(h) = \frac{GM}{(R+d)^2} $To find the ratio $\frac{g_i(d)}{g_o(d)}$:$ \frac{g_i(d)}{g_o(d)} = \frac{(R-d)}{R^3}(R+d)^2 $Simplifying further:$ = \frac{(R^2 - d^2)(R+d)}{R^3} = \frac{R^3 + R^2d - Rd^2 - d^3}{R^3} $This reduces to:$ = 1 + \frac{d}{R} - \frac{d^2}{R^2} - \frac{d^3}{R^3} $Let $\frac{d}{R} = t$, so:$ \frac{g_i(d)}{g_o(d)}= 1 + t - t^2 - t^3 $For critical points:$ \frac{d}{dt}\left(\frac{g_i(d)}{g_o(d)}\right) = 1 - 2t - 3t^2 = 0 $Factoring gives:$ 3t^2 + 3t - t - 1 = (3t-1)(t+1) $This implies critical points at:$ t = \frac{1}{3}, \quad t = -1 $Given that the distance cannot be negative, we dismiss $ t = -1 $. Therefore, the maximum value of the ratio $\frac{g_i(d)}{g_o(d)}$ occurs at:$ t = \frac{d}{R} = \frac{1}{3} $Visually, this behavior can be represented by the graph shown below:

Two spherical bodies A and B each of mass $M$ and radius $R$ are located such that their centers are...

Let us first understand all the given information.$$ v_0=2 v_{\min }, $$Here $v_{\min }$ would mean the velocity to required to reach the midpoint of $A$ and $B$ as then $C$ would have escaped the gravitational field of $A$.Now to obtain the expression, we need to use the conservation of energy to find $v_x$ :$$ \frac{1}{2} m\left(v_0\right)^2+\left(\frac{-G M m}{R}\right)+\left(\frac{-G M m}{3 R}\right)=\frac{1}{2} m\left(v_x\right)^2+\left(\frac{-G M m}{x}\right)+\left(\frac{-G M m}{4 R-x}\right) $$But before solving:Notice that $v_0$ is not present in any of the options so we need to replace $v_0$ with $2 v_{\text {min }}$ $v_{\min }$ would also be found by conservation of energy (by equation initial energy to energy at midpoint):$$ \frac{1}{2} m v_{\min }+\left(\frac{-G M m}{R}\right)+\left(\frac{-G M m}{3 R}\right)=0+\left(\frac{-G M m}{2 R}\right)+\left(\frac{-G M m}{2 R}\right) $$(velocity at mid-point will be 0 )$$ \begin{gathered} \frac{1}{2} m v_{\min }^2=\frac{G M m}{3 R} \\ \Rightarrow v_{\min }^2=\frac{2 G M}{3 R} \\ \Rightarrow v_0^2=\left(2 v_{\min }\right)^2=4 \times v_{\min }^2=\frac{8 G M}{3 R} \end{gathered} $$Now, on substituting:$$ \frac{1}{2} m\left(\frac{8 G M}{3 R}\right)+\left(\frac{-G M m}{R}\right)+\left(\frac{-G M m}{3 R}\right)=\frac{1}{2} m\left(v_x\right)^2+\left(\frac{-G M m}{x}\right)+\left(\frac{-G M m}{4 R-x}\right) $$Cancelling out $m$ :$$ \left(\frac{4 G M}{3 R}\right)+\left(\frac{-G M}{R}\right)+\left(\frac{-G M}{3 R}\right)=\frac{1}{2}\left(v_x\right)^2+\left(\frac{-G M}{x}\right)+\left(\frac{-G M}{4 R-x}\right) $$$$ \begin{aligned} &\text { LHS=0: }\\ &\begin{gathered} \frac{1}{2}\left(v_x\right)^2+\left(\frac{-G M}{x}\right)+\left(\frac{-G M}{4 R-x}\right)=0 \\ \frac{1}{2}\left(v_x\right)^2=G M\left(\frac{1}{x}+\frac{1}{4 R-x}\right) \\ \left(v_x\right)^2=2 G M\left(\frac{4 R-x+x}{(4 R-x) x}\right) \\ v_x^2=8 G M R \frac{1}{(4 R-x) x} \\ v_x=\frac{2 \sqrt{2 G M R}}{x^{\frac{1}{2}}(4 R-x)^{\frac{1}{2}}} \end{gathered} \end{aligned} $$

Gravitation - IAT-IISER Physics PYQs

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Gravitation

High-Yield Trend

Questions
2 MCQs

Official Solution

Official Solution

Gravitation

High-Yield Trend

Questions 2 MCQs

Official Solution

Official Solution

Questions
2 MCQs