Physics Guru

When a particle is thrown up vertically from the surface of the earth (Radius R), we have $v^2-u^2=2(-g)h$, $h << R$ (neglecting air, symbols have usual meanings). However, if $h << R$ condition is not met, prove that $v^2-u^2 > 2as$. Solution Using COME, $\frac {1}{2}mu^2 - \frac {GMm}{R} = \frac {1}{2} mv^2 - \frac {GMm}{R+h}$ $\Rightarrow v^2  - u^2 = 2GM ( \frac {1}{R+h} - \frac {1}{R} )$ $\Rightarrow v^2  - u^2 = 2gR^2 \frac {-h}{R(R+h)}=2gR \frac {-h}{R+h}$ $\Rightarrow v^2  - u^2 = 2(-g)h \frac {1}{1+\frac {h}{R}} > 2(-g)h$ $\Rightarrow v^2  - u^2 > 2as$; $a=-g$, $s=h$ Since $g$ falls with height, the speed falls gradually. So, it is obvious that $v^2$ will be more than what we have when $g$ is constant. The animation below shows two particles, launched at the exact same moment, with the exact same initial speed. Left particle: Constant g world Right particle: g decreases with height

Two sources of sound vibrate at the same wavelength \(\lambda\) and in phase. The setup is similar to the double slit experiment of light with one major variation, that is \(d = 2D\) (in YDSE, usually \(D \gg d\)). Will fringe width be $\frac{\lambda D}{d}$ or \(\frac{\lambda}{2}\) for \(y \ll D\)? Given that symbols have their usual meaning. Comprehensive Solution & Analysis At first glance, this question looks like a simple plug-and-chug problem. If you substitute \(d = 2D\) into the standard Young's Double-Slit Experiment (YDSE) fringe width formula \(w = \frac{\lambda D}{d}\), you get: \(w = \frac{\lambda D}{2D} = \frac{\lambda}{2}\) Algebraically, the options \(\lambda \cdot \frac{D}{d}\) and \(\frac{\lambda}{2}\) represent the exact same value. However, neither of them is correct for this setup. Here is why and how to solve it properly. 1. Why the Standard Formula Fails Here The standard formula \...

The Question Consider an object inside a pond full of water. At the critical angle C , the refracted ray just grazes the water-air interface. If the angle of incidence i > C , the light undergoes total internal reflection. Is the angle that the ray leaving the water-air interface makes with the upward normal a discontinuous function at i = C ? The Answer When the light grazes the water-air interface, the angle that the refracted ray makes with the upward normal is 90° . However, when the angle of incidence i is slightly greater than C , refraction ceases to occur and the light undergoes total internal reflection. The reflected ray then makes an angle of 180° − i with the upward normal. Thus, as i approaches C from below, the angle approaches 90°, whereas as i approaches C from above, it approaches 180° − C . Since these two limiting values are unequal, the angle that the ray leaving the water-air interface makes with the upw...

PHYSICS · RELATIVE MOTION The Stuntman, the Skis and the Rocket Launcher A fascinating problem in relative velocity Imagine a stuntman gliding effortlessly on skis through a frictionless region. He is carrying a rocket launcher and moving horizontally at a speed of v m/s. Suddenly, he fires a rocket with a muzzle speed of 4v . Here comes the interesting part: despite firing the rocket, the stuntman's velocity remains unchanged! THE CHALLENGE At what angle above the horizontal, measured relative to the ground, was the rocket launched? (Ignore air resistance.) Understanding the physics The key to solving this problem is to distinguish between the velocity of the rocket relative to the launcher and its velocity relative to the ground. Since the stuntman's velocity remains unchanged after firin...

Prove that, $[x] + \left[ {x + \frac{1}{n}} \right] + \left[ {x + \frac{2}{n}} \right] +$ ............... $+ \left[ {x + \frac{{n - 1}}{n}} \right] = [nx]$ Where, [ ] denotes greatest integer function. Solution Let, $x=I+f$ where I is the greatest integer part of $x$ and $f$ is the fractional part of $x$. We know that,  [{Integer + (.....)}] = Integer + [(.....)] $LHS = [I + f] + \left[ {I + f + \frac{1}{n}} \right] + \left[ {I + f + \frac{2}{n}} \right] +$ ............... $+ \left[ {I + f + \frac{{n - 1}}{n}} \right]$ $ = nI + [f] + \left[ {f + \frac{1}{n}} \right] + \left[ {f + \frac{2}{n}} \right] + $............... $+ \left[ {f + \frac{{n - 1}}{n}} \right]$ $ = nI + [f] + \left[ {f + \frac{1}{n}} \right] + \left[ {f + \frac{2}{n}} \right] + $.........$ + \left[ {f + \frac{p}{n}} \right] +$ ............$ + \left[ {f + \frac{{n - 1}}{n}} \right]$, where $p$ is min. value such that $\left[ {f + \frac{p}{n}} \right] = 1$ $ = nI + 0 + 0 + 0 + ........p$ times $+ \left[ {f + \frac{p}...

Evaluate, $I=\int\limits_{ - \infty }^\infty  {{e^{ - {x^2}}}dx}$ Solution $I=\int\limits_{ - \infty }^\infty  {{e^{ - {x^2}}}dx}=\int\limits_{ - \infty }^\infty  {{e^{ - {y^2}}}dy}$ ${I^2} = \left( {\int_{ - \infty }^\infty  {{e^{ - {x^2}}}} {\mkern 1mu} dx} \right)\left( {\int_{ - \infty }^\infty  {{e^{ - {y^2}}}} {\mkern 1mu} dy} \right) = \int_{ - \infty }^\infty  {\int_{ - \infty }^\infty  {{e^{ - ({x^2} + {y^2})}}} } {\mkern 1mu} dx{\mkern 1mu} dy$ Changing to polar co-ordinates, we have ${I^2} = \int_0^{2\pi } {\int_0^\infty  {{e^{ - {r^2}}}} } r{\mkern 1mu} dr{\mkern 1mu} d\theta  = \int_0^{2\pi } d \theta \int_0^\infty  r {e^{ - {r^2}}}{\mkern 1mu} dr$ [${x^2} + {y^2} = {r^2},dx{\mkern 1mu} dy = r{\mkern 1mu} dr{\mkern 1mu} d\theta $. The entire plane corresponds to $0\le r<\infty$ and $0\le\theta\le2\pi$.] Let $r^2=u$ ${I^2} = 2\pi \cdot\frac{1}{2}\int_0^\infty  {{e^{ - u}}} {\mkern 1mu} du = \pi \left[ { - {e^{ - u}}} ...

A satellite providing internet services revolves around Earth in a circular orbit at an altitude of $750km$. A ground station directly below the satellite transmits an electromagnetic signal to it and receives a response signal immediately after the satellite receives the original signal. Assume that the satellite is directly overhead during the communication, the signal travels at the speed of light, and processing delays and atmospheric effects are negligible. Calculate the minimum time taken by a signal to travel from the ground station to the satellite and back to the ground station. Compare this with a typical geostationary satellite explaining the justification behind using LEO in place of geostationary satellite taking into account the fact that TVs do operate well without LEOs. Solution Since the satellite is directly overhead, the signal travels vertically to the satellite and back to the ground station. $t_{LEO} = \frac {2\times 750 \times 10^3}{3.0 \times 10^8} = 5.0 ms$ A t...