Rocket motion along a curved path

Classical Mechanics Level 3

A rocket of mass $m_0$ moves in the absence of external forces with a constant velocity $\vec v_0$ . An engine ejecting a gas jet with a constant velocity $\vec u$ relative to the rocket at right angles to the direction of motion of the rocket is switched on. The engine is shut down when the mass of the rocket is $m$ . Through what angle (in degrees) did the direction of motion of the rocket deviate during this time?

Note : $m_0=1000$ kg., $m=950$ kg., $|\vec u|=4000$ m/s., $|\vec v_0|=40000$ m/s.

The answer is 0.2938889.

2 solutions

Aryan Sanghi
Aug 17, 2020

Let mass of rocket be $m$ and mass of fuel released be $-dm$ . Applying conservation of momentum perpendicular to rocket as no external force is there,

$dv(m + dm) = -udm$ $mdv = -udm$ $-\frac{dm}{m} = \frac{dv}{u}$ $\int_{m_o}^m-\frac{dm}{m} = \int_0^v\frac{dv}{u}$ $\log_e\bigg(\frac{m_o}{m}\bigg) = \frac{v}{u}$ $v = u\cdot \log_e\bigg(\frac{m_o}{m}\bigg)$

Putting $m_o = 1000, m = 950, u = 4000$ , we get

$v = (4000)\cdot \log_e\bigg(\frac{1000}{950}\bigg)$

$\boxed{v = 205.1m/s}$

Now,

$\tan \theta = \frac{v}{v_o}$ $\theta = \tan^{-1}\frac{205.111}{40000}$ $\color{#3D99F6}{\boxed{\theta \approx 0.293°}}$

Even though you got the answer luckily, your method of solution is wrong. See below on how to solve it :

The equation of motion of the system is

$u\dfrac{dm}{dt}=-\dfrac {mv^2}{\rho}$ ,

Where $m, u, v, \rho$ are the mass of the rocket, magnitudes of ejection velocity of the gas relative to the rocket, velocity of the rocket and radius of curvature of the path followed by the rocket respectively.

Now, $\rho =\dfrac {ds}{dα}=v\times \dfrac {dt}{dα}$

where $ds$ is the length of the arc that subtends an angle $dα$ at the centre of curvature of the path at the point considered.

So, $\dfrac {dm}{m}=-\dfrac vu dα$

Integrating this within proper limits ( $m$ from $m_0$ to $m$ , $α$ from $0$ to $\theta$ ) we get

$\ln \dfrac {m}{m_0}=-\dfrac vu \theta$

$\implies \theta =\dfrac uv \ln \left (\dfrac {m_0}{m}\right )$

Substituting values we get the result.

A Former Brilliant Member - 9 months, 4 weeks ago

But sir what is wrong in my solution? @Foolish Learner

Aryan Sanghi - 9 months, 4 weeks ago

First, it is told that $v=|\vec v|$ is constant, equal to the magnitude of initial velocity. Second, the angle between the initial and any instantaneous velocity vectors doesn't change linearly. You better take $m=1$ kg. when the engine is shut down and find the angle. You will get your fault. You got the correct answer because the angle was small, and for small angles $α$ and $\tan α$ are approximately equal.

A Former Brilliant Member - 9 months, 4 weeks ago

@A Former Brilliant Member – Ohk, Now I understood sir. Thanks a lot for explaining me sir. :)

Aryan Sanghi - 9 months, 4 weeks ago

@Steven Chase sir has also used same method as mine. I understood your method sir, but it's long. :)

Aryan Sanghi - 9 months, 4 weeks ago

Don't see what the others do. Build a confidence in yourself. Think independently and logically.

A Former Brilliant Member - 9 months, 4 weeks ago

@A Former Brilliant Member – Yes sir, I'll follow this advice. :)

Aryan Sanghi - 9 months, 4 weeks ago

Steven Chase
Aug 17, 2020

Nice problem. Commented solution code is attached:

import math

m0 = 1000.0
mf = 950.0
vr = 4000.0     # relative exhaust velocity
v0 = 40000.0

md = 1000.0     # mass ejection rate

dt = 10.0**(-8.0)  # time step

######################################

# Initialize simulation

t = 0.0

m = m0

vx = v0
vy = 0.0

ax = 0.0
ay = 0.0

while m >= mf:        # run simulation until stop condition

    vx = vx + ax*dt  # numerical integration
    vy = vy + ay*dt

    m = m - md*dt    # update mass

    v = math.hypot(vx,vy)

    ux = -vy/v   # unit vector in force direction
    uy = vx/v

    F = md*vr    # exhaust force magnitude

    Fx = F*ux    # exhaust force vector
    Fy = F*uy

    ax = Fx/m    # acceleration
    ay = Fy/m

    t = t + dt

######################################

# print results

print dt
print ""
print vx
print vy
print ""

theta = math.atan(vy/vx)

print (180.0*theta/math.pi)

######################################

#>>> 
#1e-06

#39999.4738107
#205.172383155

#0.293889079313
#>>> ================================ RESTART ================================
#>>> 
#1e-07

#39999.4738017
#205.172288394

#0.293888943645
#>>> ================================ RESTART ================================
#>>> 
#1e-08

#39999.4738008
#205.172278906

#0.293888930061
#>>>

Where did you get the value of mass eject rate (md) ?

Hosam Hajjir - 9 months, 4 weeks ago

I just picked an arbitrary value. The answer is independent of the particular value chosen

Steven Chase - 9 months, 4 weeks ago

You mean to say deviation can never be more than $90\degree$ ? Take $m_0=20\times 10^{10}$ kg., $m=950$ kg., and run your programme.

A Former Brilliant Member - 9 months, 4 weeks ago

@Foolish Learner Fuck off.

Talulah Riley - 8 months, 3 weeks ago

Rocket motion along a curved path

The answer is 0.2938889.

2 solutions

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