Rocket propulsion - Differential equations

  • Thread starter Locoism
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Homework Statement




Suppose a rocket is launched from the surface of the earth with initial velocity
[itex] v_0 = \sqrt(2gR) [/itex], the escape velocity.

a) Find an expression for the velocity in terms of the distance x from the surface of the earth (ignore air resistance)

b) Find the time required for the rocket to go 240,000 miles. Assume R = 4000 miles, and g = 78,545 miles/h2


The Attempt at a Solution



So I figure we use [itex] \frac{dv}{dt} = -\frac{mG}{(R+x)^2} [/itex] and multiply by [itex]\frac{dt}{dx} = \frac{1}{v}[/itex] to get

[itex] \frac{dt}{dx} \frac{dv}{dt} = \frac{dv}{dx} = -\frac{mG}{v (R+x)^2} [/itex]

which gives [itex] v(x) = \sqrt(\frac{2mG}{R+x}) + C [/itex]

Using [itex] v(0) = \sqrt(2gR) [/itex] gives

[itex] C = \sqrt(2gR) - \sqrt(\frac{2mG}{R}) [/itex]


I don't know where to go from here....
 

Answers and Replies

  • #2
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g, G, m and R are related. Use the relationship to eliminate m and G from your formula for v(x). That should answer (a).

For (b), you just need to integrate (a).
 
  • #3
HallsofIvy
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Problems like this really annoy me! The whole point of a rocket is that, unlike simply throwing a rock or ball in the air, is that a rocket engine stays on during its flight. In this problem we are apparently to assume that is not true.
 
  • #4
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Problems like this really annoy me! The whole point of a rocket is that, unlike simply throwing a rock or ball in the air, is that a rocket engine stays on during its flight. In this problem we are apparently to assume that is not true.

Lol. But what if we assume the rocket is very big ball thrown by a really strong dude? o_O


As for the actual math...

g, G, m and R are related. Use the relationship to eliminate m and G from your formula for v(x). That should answer (a).

For (b), you just need to integrate (a).

What? How do I do that??? It seems that whenever I integrate of differentiate they will always remain since they multiply y? The only way I can see they are related is that they appear in the same formula for y''... How can I cancel them?
 
  • #5
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mg equals the force of gravity on the surface of the Earth. What is the the force of gravity on the surface of the Earth?

edit: "m" here is not the mass of the Earth as you used it above. It is the mass of some small body, such as the rocket. I suggest you denote the mass of the Earth as M to eliminate confusion.
 

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