Question

A space probe approaches a planet and goes into a low orbit. If the orbiting probe's velocity is
vx=−v sin(ωt),
vy=v cos(ωt),
v = 7.77 103 m/s,
ω= 1.20 10-3 radians/s,
what is the average magnitude of acceleration of the probe if it remains in orbit for 24 minutes?

Answer 1

here given that the velocity of the probe is

$v_x = - v sin\omega t$

$v_y = v cos\omega t$

now at initial position where t = 0

$v_{xi} = - v sin0 = 0$

$v_{yi} = v cos 0 = v$

Now after t = 24 minutes we need to find final components of velocity

$v_{xf} = - v sin(1.20* 10^{-3} * 24*60) = -v sin(1.728)$

$v_{yf} =  v cos(1.20* 10^{-3} * 24*60) = v cos(1.728)$

now as we know that acceleration is given as

$a = \frac{v_f - v_i}{t}$

Now for x direction of motion

$a_x = \frac{v_{xf} - v_{xi}}{t}$

$a_x = \frac{- v sin(1.728) - 0}{24*60}$

$a_x = \frac{-7.77*10^3 * 0.99}{24*60}$

$a_x = -5.33 m/s^2$

Now for y direction of motion

$a_y = \frac{v_{yf} - v_{yi}}{t}$

$a_y = \frac{ v cos(1.728) - v}{24*60}$

$a_y = \frac{7.77*10^3 * (-0.16) - 7.77 * 10^3}{24*60}$

$a_y = -6.24 m/s^2$

now in order to find the magnitude of acceleration we can say

$a = \sqrt{a_x^2 + a_y^2}$

$a = \sqrt{5.33^2 + 6.24^2} = 8.2 m/s^2$