Question

A satellite is moving in an elliptical orbit about a planet. The equation of it’s planar orbit is 3x^2+4y^2=20. If the velocity of the satellite in the y direction is 10 when the y coordinate of the satellite is 2, what is the velocity in the x direction at that time?

Answer 1

Answer:

Explanation:

Given

Equation of Planet orbit

$3x^2+4y^2=20---1$

also $v_y=10\ unit/s$

when $y=2\ unit$

Substitute the value of y in 1

$3x^2+4(2)^2=20$

$x^2=\frac{4}{3}$

$x=\frac{2}{\sqrt{3}}$

differentiate 1 to get velocity in x direction

$3\times 2\times x\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+4\times 2\times y\cdot \frac{\mathrm{d} y}{\mathrm{d} t}=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=v_x$

$\frac{\mathrm{d} y}{\mathrm{d} t}=v_y$

$v_x=-\frac{4}{3}\times \frac{y}{x}\times v_y$

$v_x=-\frac{4}{3}\times \frac{2\sqrt{3}}{2}\times 10$

$v_x=-23.094\ unit/s$