Problem set 5

1. A planet moves around the sun in an elliptic orbit with length of major axis equal to 1.524 times that of the Earth. The time of revolution of the planet about the Sun is
1. 1 year
2. 10.24 year
3. 0.5315 year
4. 1.8814 year
According to Kepler's third law, the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. Hence, period of revolution of a planet around the Sun in an elliptic orbit is given by $$T^2=K_sa^3$$, where $K_s=2.97\times10^{-19}s^2/m^3$. $$T^2_{earth}=K_sa^3_{earth}$$ $$T^2_{planet}=K_sa^3_{planet}$$ $$\frac{T^2_{planet}}{T^2_{earth}}=\frac{a^3_{planet}}{a^3_{earth}}$$ $$\frac{T^2_{planet}}{1}=\frac{\left(1.524a_{earth}\right)^3}{a^3_{earth}}$$ $$T_{planet}=\left(1.524\right)^{3/2}=1.8814~ years$$
2. A particle is released from a large height $h$, at a location with latitude $\lambda$. At the time of striking the ground, the horizontal deflection that occurs due to Coriolis force, is proportional to
1. $\sin{\lambda}$
2. $\cos{\lambda}$
3. $\sec{\lambda}$
4. $cosec\lambda$
The horizontal deflection is given by $$y(h)=\frac{2h\omega\sin\theta}{3}\sqrt{\frac{2h}{g}}$$ where $\theta$ is colatitude. If $\lambda$ is latitude then $\sin\theta=cos\lambda$ $$y(h)=\frac{2h\omega\cos\lambda}{3}\sqrt{\frac{2h}{g}}$$
3. The centre of the circle $\bar z z+(2+3i)\bar z+(2-3i)z+1=0$ is
   1. (2,3)
   2. (3,2)
   3. (-2,-3)
   4. (4,0)
   Let $z=x+iy$ \begin{align*} \therefore \bar z z+(2+3i)\bar z+(2-3i)z+1\\ =(x-iy)(x+iy)+(2+3i)(x-iy)\\+(2-3i)(x+iy)+1\\ =x^2+y^2+2x-2iy+3ix+3y\\+2x+2iy-3ix+3y+1\\ =x^2+y^2+4x+6y+1 \end{align*} $$\therefore\bar z z+(2+3i)\bar z+(2-3i)z+1=0$$ gives $$(x+2)^2+(y+3)^2-12=0$$ $(x+2)^2+(y+3)^2=12$ is equation of circle with centre $(-2,-3)$ and radius $\sqrt{12}$
4. Fourier transform of the function $f(x)=exp(-|x|)$ is
   1. $\frac{1}{\sqrt{2\pi}}\left[\frac{2}{1+k^2}\right]$
   2. $0$
   3. $\frac{1}{\sqrt{\pi}}\left[\frac{1}{k^2}\right]$
   4. $\frac{1}{\sqrt{2\pi}}\left[\frac{1}{1-k^2}\right]$
   $|x|=x$ if $x>0$ and $|x|=-x$ if $x<0$. Fourier transform of $f(x)$ is given by $$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x)dx$$ \begin{align*} F(k)={\scriptstyle\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}e^{-ikx}e^xdx+\int_{0}^{\infty}e^{-ikx}e^{-x}dx\right]} \end{align*} $$F(k)={\scriptstyle\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}e^{x(1-ik)}dx+\int_{0}^{\infty}e^{-x(1+ik)}dx\right]}$$ $$F(k)=\frac{1}{\sqrt{2\pi}}\left[\frac{2}{1+k^2}\right]$$
5. The area of the triangle whose base is given by $\bar a=5\hat i-3\hat j+4\hat k$ and $\bar b=\hat j-\hat k$ is another side is :
   1. $\sqrt{50}/2$
   2. $\sqrt{61}/2$
   3. $\sqrt{14}/2$
   4. $\sqrt{51}/2$
   \begin{align*} &\text{Area of triangle }=\frac{1}{2}|\bar a\times\bar b|\\ &=\frac{1}{2}|-\hat i+5\hat j+5\hat k|=\sqrt{51}/2 \end{align*}