A spaceship, mass 1.00 times 10^3 kg, is moving with constant velocity of magnit

Question

A spaceship, mass 1.00 times 10^3 kg, is moving with constant velocity of magnitude 100 ms-1 in the y- direction in a region of space where there is negligible gravitational field. To change course, the pilot fire a thruster rocket which applies a constant force F, in the x- direction as shown in Fig for 10.0 seconds. The rocket is then switched off, and the speed of the spacecraft is 112ms-1. (You may assume that the spacecraft does not rotate.) By how much is the kinetic energy of the craft increased by firing the thruster rocket? What is the magnitude of F? Calculate the components of the total displacement of the spacecraft from its position at the beginning of the 10.0 second interval to its position at the end of the 10.0 second interval. Without doing any further calculation, sketch the track of the spacecraft, showing the motion during the period that the thruster rocket is firing and the path the craft will take after the rocket is turned off. (Mark distance and velocities on your sketch where known.)Calculate the velocity of spacecraft at time t = 11.0 seconds(i.e. 1 second after rocket it switched off), and include this on your sketch. On retuning towards the Earth, the pilot wishes to place his craft in an orbit that is 250 km above the surface of the Earth . (Radius of Earth about 6400km) What speed must the spacecraft have to be stable in this orbit? (You may assume that the acceleration due to gravity, g, is 9.10ms at 250 km above the Earth's surface.

Explanation / Answer

1)Change in kinetic energy= Final kinetic energy- inital kinetic energy

Final KE= 0.5*1000*(112^2)= 6272000 J

Inital KE = 0.5*1000*(100^2) = 5000000 J

Difference= 6272000-5000000= 1272000J= 1.272*(10^6)J

2) Speed in X direction= sqrt (final speed^2 - initial speed^2)

= sqrt(112^2 - 100^2)= 50.44 m/s

v = acceleration*time
v = 50.44m/s, t=10s
a = 50.44/10 = 5.044 m/s^2

F = ma = 1000*5.044 = 5044 N

3) x = 1/2 a t^2 = 0.5*5.044*10^2=252.2 m

y = vt = 100*10 = 1000 m

vx = 5.044*11 = 55.48 m/s
and vy = 100 m/s

4)
to be stable
G M m/r^2 = mv^2/r

G M/R^2 = g
so

g R^2/r^2 = mv^2/r

9.81*6400E3^2/(6400E3+250.0E3) = v^2

v=7773 m/s

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