Question

Around midcourse, a velocity adjustment is performed to eliminate the small errors introduced when departing from Earth orbit. This adjustment is performed using one of the onboard thrusters. At the location where the adjustment is made, the velocity is V = 26,300 m/s and should be V = 27,000 m/s. Knowing that the thruster used for the maneuver generates a thrust F = 7,740 N, determine how long, in minutes, it should be turned on to adjust the velocity. The mass of the spacecraft is 2,500 kg

Answer 1

\[ F_{thrust} = ma_{thrust} \\ v_f = v_i + a_{thrust}t\]
Make sure to convert the time from second to minutes :)

Answer 2

how exactly do I set everything up?

Answer 3

Your velocity is in the +x direction, as is your acceleration - thrust in this context is basically like saying the spaceship is being pulled by a rope to be sped up; the resultant force is in the positive x direction. There's no gravity to apply or anything, so your summation of forces is just
\[ \sum F = F_{thrust} = ma\]
\[ \sum F = ma_{thrust} = ma\]

So your acceleration is
\[ a_{thrust} = \frac{F_{thrust}}{m}\]

Your kinematic equation is an acceleration definition
\[ \frac{\Delta v}{\Delta t}=a\]
except that you already know the acceleration and are solving for t.
\[ \frac{ v-v_o}{t - t_o }=a_{thrust}\]
\[ \frac{ v-v_o}{a_{thrust} }=t - t_o\]

\[ t_o = 0
\\ t = ?
\\ v_o = 26,300 m/s
\\ v = 27,000 m/s
\\ a_{thrust} = \frac{F_{thrust}}{m} = \frac{7740N}{2500kg} \]