Question

1. A certain planet has an average surface area that is 2.25% larger than the earth and a gravitational acceleration that is three fifth times than that of the earth’s gravitational acceleration. What is the velocity of escape on that planet? Ans. 8728.33 m/s
2. Find the isogonal trajectories of the family of curves x²+y²=c if θ=45°. Ans. ln(x²+y²)+2 arctan (x/y) =k

Answer 1

1.

For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula

$$v_e=\sqrt{\dfrac{2GM}{r}}$$

where $G$ is the universal gravitational constant, $M$  is the mass of the body to be escaped from, and $r$ is the distance from the center of mass of the body to the object

An alternative expression for the escape velocity $v_e$ is

$$v_e=\sqrt{2gr}$$

where $r$ is the distance between the center of the body and the point at which escape velocity is being calculated and $g$ is the gravitational acceleration at that distance.

Then

$$v_e=\sqrt{2(\dfrac{3}{5})(9.81m/s^2)(\sqrt{1.0225}(6.4\times10^6m)}$$

$$\approx8728.33m/s$$

2.

$$x^2+y^2=c$$

Differentiate both sides with respect to $x$

$$2x+2y\dfrac{dy}{dx}=0$$

$$x+y\dfrac{dy}{dx}=0$$

Replace $\dfrac{dy}{dx}$ by $\dfrac{\dfrac{dy}{dx}-\tan 45\degree}{1+\dfrac{dy}{dx}\tan 45\degree}=\dfrac{\dfrac{dy}{dx}-1}{1+\dfrac{dy}{dx}(1)}$ we get

$$x+y\dfrac{\dfrac{dy}{dx}-1}{1+\dfrac{dy}{dx}}=0$$

$$x(1+\dfrac{dy}{dx})+y(\dfrac{dy}{dx}-1)=0$$

$$\dfrac{dy}{dx}=\dfrac{y-x}{y+x}$$

Put $y=ux, \dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

$$x\dfrac{du}{dx}+u=\dfrac{ux-x}{ux+x}$$

$$x\dfrac{du}{dx}=\dfrac{u-1-u^2-u}{u+1}$$

$$\dfrac{u+1}{u^2+1}du=-\dfrac{dx}{x}$$

Integrate

$$\int\dfrac{u+1}{u^2+1}du=-\int\dfrac{dx}{x}$$

$$\dfrac{1}{2}\ln(u^2+1)+\tan^{-1}u=-\ln|x|+\dfrac{1}{2}k$$

$$\ln(x^2u^2+x^2)+2\tan^{-1}u=k$$

$$\ln(y^2+x^2)+2\tan^{-1}(y/x)=k$$

which is the required trajectories of the given family of curves.