Rocket propulsion - Differential equations

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Homework Help Overview

The problem involves a rocket launched from the Earth's surface, focusing on deriving expressions for velocity and time using differential equations. The context includes gravitational forces and the dynamics of rocket propulsion, specifically under the assumption of no air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of differential equations to express velocity in terms of distance and the relationship between various physical constants. There is also a focus on integrating the derived expression for velocity to find time.

Discussion Status

Some participants have provided guidance on eliminating variables from the velocity expression and integrating it for the time calculation. Others express frustration with the assumptions made in the problem, particularly regarding the rocket's propulsion.

Contextual Notes

There is a noted confusion regarding the definitions of variables, particularly the mass of the rocket versus the mass of the Earth, which may affect the clarity of the discussion.

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Homework Statement




Suppose a rocket is launched from the surface of the Earth with initial velocity
v_0 = \sqrt(2gR), 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 \frac{dv}{dt} = -\frac{mG}{(R+x)^2} and multiply by \frac{dt}{dx} = \frac{1}{v} to get

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

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

Using v(0) = \sqrt(2gR) gives

C = \sqrt(2gR) - \sqrt(\frac{2mG}{R})


I don't know where to go from here...
 
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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).
 
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.
 
HallsofIvy said:
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...

voko said:
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?
 
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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