Least Distance"

Classical Mechanics Level 3

A satellite is orbiting around Earth in a circular orbit of radius $r.$ A particle of mass $m$ is projected from the satellite in a forward direction with velocity $v = \sqrt {\frac 23} v_0$ , where $v_0$ is the orbital velocity of the satellite. During subsequent motion of the particle, its minimum distance from the center of Earth is given by

$d_\text{min}=\dfrac {ar}{b},$

where $a$ and $b$ are coprime positive integers.

Find $a + b.$

The answer is 3.

1 solution

A Former Brilliant Member
Dec 21, 2018

As the velocity of particle is less than the orbital velocity of satellite , the particle goes in an elliptical orbit of semi-major axis less than $r.$

Let $r_1$ be the minimum distance and $v_1$ be the velocity of the particle at that position, $M_e$ is mass of Earth, then

$m\times \sqrt{\dfrac 23} v_0 r = mv_1 r_1$

$v_1 r_1 = \sqrt{\dfrac 23} v_0 r$

From Energy Conservation,

$\dfrac {m \times \dfrac 23 v_0^2 }{2} - \dfrac {GM_em}{r} = \dfrac {mv_1^2}{2} - \dfrac {GM_em}{r_1}$

Here, $v_0 = \sqrt{\dfrac {GM_e}{r}}$

Solving we get, $r_1 = d_{min} = \dfrac r2 = \dfrac {1\times r}{2} = \dfrac {ar}{b}$

Hence , $a + b = 1 + 2 = \boxed {3}$

Least Distance"

The answer is 3.

1 solution

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