Problem Solving Assignment

1. Rocket launching: In this problem we are looking

at some of the aspects that need to be considered

in designing and describing the launching a single-

stage rocket with a payload. In what follows we will

be ignoring any eects of cross-winds, rotation of

Earth or air resistance.

We begin by assuming that the rocket is launched

from Earth's surface in such a way that it follows a

simple vertical trajectory. As a single-stage rocket,

immediately after launch there will be a period of

rapid acceleration as the fuel consumed provides

thrust for the upward motion. One of the complex

issues to be considered is that as the fuel is used, the

mass of the rocket (i.e. fuel + engine + payload)

decreases, and hence the acceleration will increase

for the same amount of thrust. The period of time

over which the rocket engine is ring is known as the

`boost phase' of the trajectory, hence when the fuel

runs out, the `boost phase' is at an end.

Representing the rocket's total mass with the un-

known function m(t), if we assume that one of the

design parameters for the rocket engine is to set a

constant fuel burn rate, then we will have the rela-

tionship dm=dt = ô€€€b where b is the fuel burn rate

constant measured in kg sô€€€1. The initial total mass

of the rocket can be represented by the constant m0,

and we can assume that the fuel is a signicant frac-

tion of this mass. In this case, the quantity of fuel

can again be a design parameter, so we will take the

mass of fuel to be X m0, where 0 < X < 1.

If we further assume that the `boost phase' is short

enough, and we assign the function h(t) to be the

rocket's height above ground-level, then we can de-

scribe the acceleration experienced by the rocket

during this period as

d2h

dt2 =

b

m0 ô€€€ b t

ô€€€ g ;

where g is the acceleration due to gravity, and b is

the thrust provided by the rocket engine, being an

experimentally determined constant. If we launch

the rocket from a stationary position, we can jus-

tiably include the initial conditions h(0) = 0 and

dh=dt = 0 when t = 0.

(a) Show that the expression for the height h(t) of

the rocket when it just runs out of fuel is given

by

m0

b

(1 ô€€€ X) ln(1 ô€€€ X) + X

1 ô€€€

g

2b

Xm0

:

(b) What is the velocity of the rocket at the end of

the `boost phase'?

(c) Astronauts aboard the `vomit comet' or its

current equivalent1, experience a rapid loss of

weight as the acceleration due to gravity is

counteracted by the acceleration provided by

the aircraft. The rate of weight-loss expe-

rienced is equivalent to a rate of decrease in

the acceleration due to gravity of the order of

ô€€€10=15 msô€€€2=s = ô€€€2=3 msô€€€3.

For our rocket and its payload, the acceleration

due to gravity decreases as the rocket gets fur-

ther from the launch site - still assuming a ver-

tical trajectory. If we take the radius of Earth

to be R0 = 6:38 106 m, then the acceleration

due to gravity goes like

ag =

MG

(h(t) + R0)2 ;

where M and G are constants2 and as before,

h(t) is the rocket's height above Earth's surface

as a function of time. For an object of constant

mass, but changing height, the rate of weight-

loss experienced would be given by d(ag)=dt.

Using the information gained previously for the

height and velocity of the rocket in the `boost

phase', nd an expression for the rate of weight-

loss experienced by the rocket and it's payload

as it continues to climb just at the end of the

boost phase. You should use the operating pa-

rameters of the Shahab 3 missile employed by

Iran (see notes3 4), which has an initial mass

of 16000 kg of which 90% can be considered as

fuel. Furthermore the Shahab 3 has a boost

phase of 95 s by which time it reaches a height

of 75 km. How does this rate of weight-loss

compare with that experienced in the `vomit

comet'?

1. Rocket launching: In this problem we are looking

at some of the aspects that need to be considered

in designing and describing the launching a single-

stage rocket with a payload. In what follows we will

be ignoring any eects of cross-winds, rotation of

Earth or air resistance.

We begin by assuming that the rocket is launched

from Earth's surface in such a way that it follows a

simple vertical trajectory. As a single-stage rocket,

immediately after launch there will be a period of

rapid acceleration as the fuel consumed provides

thrust for the upward motion. One of the complex

issues to be considered is that as the fuel is used, the

mass of the rocket (i.e. fuel + engine + payload)

decreases, and hence the acceleration will increase

for the same amount of thrust. The period of time

over which the rocket engine is ring is known as the

`boost phase' of the trajectory, hence when the fuel

runs out, the `boost phase' is at an end.

Representing the rocket's total mass with the un-

known function m(t), if we assume that one of the

design parameters for the rocket engine is to set a

constant fuel burn rate, then we will have the rela-

tionship dm=dt = ô€€€b where b is the fuel burn rate

constant measured in kg sô€€€1. The initial total mass

of the rocket can be represented by the constant m0,

and we can assume that the fuel is a signicant frac-

tion of this mass. In this case, the quantity of fuel

can again be a design parameter, so we will take the

mass of fuel to be X m0, where 0 < X < 1.

If we further assume that the `boost phase' is short

enough, and we assign the function h(t) to be the

rocket's height above ground-level, then we can de-

scribe the acceleration experienced by the rocket

during this period as

d2h

dt2 =

b

m0 ô€€€ b t

ô€€€ g ;

where g is the acceleration due to gravity, and b is

the thrust provided by the rocket engine, being an

experimentally determined constant. If we launch

the rocket from a stationary position, we can jus-

tiably include the initial conditions h(0) = 0 and

dh=dt = 0 when t = 0.

(a) Show that the expression for the height h(t) of

the rocket when it just runs out of fuel is given

by

m0

b

(1 ô€€€ X) ln(1 ô€€€ X) + X

1 ô€€€

g

2b

Xm0

:

(b) What is the velocity of the rocket at the end of

the `boost phase'?

(c) Astronauts aboard the `vomit comet' or its

current equivalent1, experience a rapid loss of

weight as the acceleration due to gravity is

counteracted by the acceleration provided by

the aircraft. The rate of weight-loss expe-

rienced is equivalent to a rate of decrease in

the acceleration due to gravity of the order of

ô€€€10=15 msô€€€2=s = ô€€€2=3 msô€€€3.

For our rocket and its payload, the acceleration

due to gravity decreases as the rocket gets fur-

ther from the launch site - still assuming a ver-

tical trajectory. If we take the radius of Earth

to be R0 = 6:38 106 m, then the acceleration

due to gravity goes like

ag =

MG

(h(t) + R0)2 ;

where M and G are constants2 and as before,

h(t) is the rocket's height above Earth's surface

as a function of time. For an object of constant

mass, but changing height, the rate of weight-

loss experienced would be given by d(ag)=dt.

Using the information gained previously for the

height and velocity of the rocket in the `boost

phase', nd an expression for the rate of weight-

loss experienced by the rocket and it's payload

as it continues to climb just at the end of the

boost phase. You should use the operating pa-

rameters of the Shahab 3 missile employed by

Iran (see notes3 4), which has an initial mass

of 16000 kg of which 90% can be considered as

fuel. Furthermore the Shahab 3 has a boost

phase of 95 s by which time it reaches a height

of 75 km. How does this rate of weight-loss

compare with that experienced in the `vomit

comet'?

### Recently Asked Questions

- In Fig. 2, a string, tied to a sinusoidal oscillator at p and running over a support at Q , is stretched by a block of mass M. Separation

- In an all boys school, the heights of the student body are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. Out of the 447

- For males in a certain town, the systolic blood pressure is normally distributed with a mean of 105 and a standard deviation of 5. What percentage of males in