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 = (v_f - v_i)/(t)`

Now for x direction of motion

`a_x = (v_(xf) - v_(xi))/(t)`

`a_x = (- v sin(1.728) - 0)/(24*60)`

`a_x = (-7.77*10^3 * 0.99)/(24*60)`

`a_x = -5.33 m/s^2`

Now for y direction of motion

`a_y = (v_(yf) - v_(yi))/(t)`

`a_y = ( v cos(1.728) - v)/(24*60)`

`a_y = (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 = √(a_x^2 + a_y^2)`

`a = √(5.33^2 + 6.24^2) = 8.2 m/s^2`