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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

what is torque
Deepak Reply
The turning effect of force is called torque.
Uzair
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Promise Reply
formula for impedance
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Victoria Reply
ohms law state that the electricity passing through a metallic conductor is directly proportional to the potential difference across its end
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folajin Reply
Anything that occupies space
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Right
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the time rate of increase in velocity is called
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acceleration
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P=F/A
Mira
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Bern
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forces acting and lying on d same plane
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How does a current follow?
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follow?
akif
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?
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AC current
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AC current follows due to changing electric field and magnetic field.
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flows
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Vineeta
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Vineeta Reply
substances having lower forbidden gap between valence band and conduction band
akif
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akif
convert 56°c to kelvin
Abubakar
How does a current follow?
Vineeta
A semiconductor is any material whose conduction lies between that of a conductor and an insulator.
AKOWUAH
what is Atom? what is molecules? what is ions?
Abubakar Reply
atoms are the smallest unit of an element which is capable of behaving as a single unit
Promise
a molecule is d smallest unit of a substances capable of independent existence and can also retain the chemical proper ties of that substance
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an ion is referred to as freely moving charged particles
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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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