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A rocket’s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket . First, the greater the exhaust velocity of the gases relative to the rocket, v e size 12{v rSub { size 8{e} } } {} , the greater the acceleration is. The practical limit for v e size 12{v rSub { size 8{e} } } {} is about 2 . 5 × 10 3 m/s size 12{2 "." 5 times "10" rSup { size 8{3} } `"m/s"} {} for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor Δ m / Δ t size 12{Δm/Δt} {} in the equation. The quantity ( Δ m / Δ t ) v e size 12{ \( Δm/Δt \) v rSub { size 8{e} } } {} , with units of newtons, is called "thrust.” The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass m size 12{m} {} of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass m size 12{m} {} decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted.

Factors affecting a rocket’s acceleration

  • The greater the exhaust velocity v e size 12{v rSub { size 8{e} } } {} of the gases relative to the rocket, the greater the acceleration.
  • The faster the rocket burns its fuel, the greater its acceleration.
  • The smaller the rocket’s mass (all other factors being the same), the greater the acceleration.

Calculating acceleration: initial acceleration of a moon launch

A Saturn V’s mass at liftoff was 2 . 80 × 10 6 kg size 12{2 "." "80" times "10" rSup { size 8{6} } `"kg"} {} , its fuel-burn rate was 1 . 40 × 10 4 kg/s size 12{1 "." "40" times "10" rSup { size 8{4} } `"kg/s"} {} , and the exhaust velocity was 2 . 40 × 10 3 m/s size 12{2 "." "40" times "10" rSup { size 8{3} } `"m/s"} {} . Calculate its initial acceleration.

Strategy

This problem is a straightforward application of the expression for acceleration because a size 12{a} {} is the unknown and all of the terms on the right side of the equation are given.

Solution

Substituting the given values into the equation for acceleration yields

a = v e m Δ m Δ t g = 2 . 40 × 10 3 m/s 2 . 80 × 10 6 kg 1 . 40 × 10 4 kg/s 9 . 80 m/s 2 = 2 . 20 m/s 2 .

Discussion

This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because m size 12{m} {} decreases while v e size 12{v rSub { size 8{e} } } {} and Δ m Δ t size 12{ { {Δm} over {Δt} } } {} remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was 3 . 36 × 10 7 N size 12{3 "." "36" times "10" rSup { size 8{7} } `N} {} .

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To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth’s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is

v = v e ln m 0 m r , size 12{v=v rSub { size 8{e} } "ln" { {m rSub { size 8{0} } } over {m rSub { size 8{r} } } } ,} {}

where ln m 0 / m r size 12{"ln"` left (m rSub { size 8{0} } /m rSub { size 8{r} } right )} {} is the natural logarithm of the ratio of the initial mass of the rocket m 0 size 12{ left (m rSub { size 8{0} } right )} {} to what is left m r size 12{ left (m rSub { size 8{r} } right )} {} after all of the fuel is exhausted. (Note that v size 12{v} {} is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth’s gravity starting from rest, given that the escape velocity from Earth is about 11 . 2 × 10 3 m/s size 12{"11" "." 2 times "10" rSup { size 8{3} } `"m/s"} {} , and assuming an exhaust velocity v e = 2 . 5 × 10 3 m/s size 12{v rSub { size 8{e} } =2 "." 5 times "10" rSup { size 8{3} } `"m/s"} {} .

ln m 0 m r = v v e = 11 . 2 × 10 3 m/s 2 . 5 × 10 3 m/s = 4 . 48 size 12{"ln" { {m rSub { size 8{0} } } over {m rSub { size 8{r} } } } = { {v} over {v rSub { size 8{e} } } } = { {"11" "." 2 times "10" rSup { size 8{3} } `"m/s"} over {2 "." 5 times "10" rSup { size 8{3} } `"m/s"} } =4 "." "48"} {}

Solving for m 0 / m r size 12{m rSub { size 8{0} } /m rSub { size 8{r} } } {} gives

Questions & Answers

how do you calculate the 5% uncertainty of 4cm?
melia Reply
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
melia Reply
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
which topic u discussing about?
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
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Chioma
hi
haider
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sean
hi
Noks
the meaning of phrase in physics
Chovwe Reply
is the meaning of phrase in physics
Chovwe
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
Gabriel Reply
how does a model differ from a theory
Friday Reply
To use the vocabulary of model theory and meta-logic, a theory is a set of sentences which can be derived from a formal model using some rule of inference (usually just modus ponens). So, for example, Number Theory is the set of sentences true about numbers. But the model is a structure together wit
Jesilda
with an iterpretation.
Jesilda
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Ridwan Reply
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
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Jack Reply
what's electromagnetic induction
Chinaza Reply
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
Ajayi Reply
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
Gabriel Reply
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
Gabriel Reply
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
Ademiye
yes it was an assignment question "^"represent raise to power pls
Gabriel
mu/y³ u>v²k² uk²/√u-vk please help me out
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An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
Imtiaz Reply
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Ademiye
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
Isaac Reply
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
Mildred Reply
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
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method of polarization
Ajayi
What is atomic number?
Makperr Reply
The number of protons in the nucleus of an atom
Deborah

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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