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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

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.077 m2, and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross
fagbeji Reply
A particle behave like a wave and we do not why?
WAQAR
what's the period of velocity 4cm/s at displacement 10cm
Andrew Reply
What is physics
LordRalph Reply
the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.
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and the word of matter is anything that have mass and occupied space
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1/v+1/u=1/f
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a man has created by who?
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What type of experimental evidence indicates that light is a wave
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double slit experiment
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ENOBONG Reply
some sort of blatherring or mambo jambo you may say
muhammad
I still don't understand what this group is all about oo
ENOBONG
no
uchenna
ufff....this associated with physics ..so u can ask questions related to all topics of physics..
muhammad
what is sound?
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what is upthrust
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what is upthrust
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Up thrust is a force
Samuel
upthrust is a upward force that acts vertical in the ground surface.
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yes rodney's answer z correct
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what is centre of gravity?
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you think the human body could produce such Force
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mirobiology
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what is specific latent heat
Omosebi Reply
the total amount of heat energy required to change the physical state of a unit mass of matter without a corresponding change in temperature.
fitzgerald
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muhammad
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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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