# Homework Help: Horizontal escape of a projectile

1. Jun 17, 2011

### kraigandrews

1. The problem statement, all variables and given/known data

The problem is attached

2. Relevant equations

L=mr^2(dr/dt)
V_escape=(GM/R)^.5
E=L^2/(2mR^2)-(GMm)/R=0

3. The attempt at a solution
after some work i got:
dr/dtheta=((r^3/R)-r^2)^1/2
dont know what to do from here.
Thanks.

#### Attached Files:

• ###### escape problem.jpg
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2. Jun 17, 2011

### Staff: Mentor

I'm not completely clear on what it is you're trying to find. Is it the minimum velocity for escape for a horizontal launch?

3. Jun 17, 2011

### kraigandrews

I'm trying to find r(theta)

4. Jun 17, 2011

### Staff: Mentor

I see. Well, you're at the point where you'll have to integrate. The equation is separable into r and θ portions.

5. Jun 17, 2011

### kraigandrews

then second part of my question i have to do is find the coordinates where r(theta) crosses the y-axis however, where i run into the problem is the integral i found doesnot cross the y-axis

6. Jun 17, 2011

### Staff: Mentor

Hmm. What did your integration and its result look like?

7. Jun 17, 2011

### kraigandrews

well seperating them it is dr/((r^3/R)-r^2)^1/2)=dtheta
i used wolframalpha and got:
what is attached and I used R= to 6.4E6m (radius of the earth)

#### Attached Files:

• ###### wolfram.gif
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8. Jun 17, 2011

### Staff: Mentor

If you keep things in symbolic form a little longer, some good things might happen

Attached is a snapshot of a Wolfram Integrator output where for the function I've pulled an r2 out of the radical before letting the integrator loose.

Thus you have:
$$\theta = 2 tan^{-1} \left( \sqrt{\frac{r}{r_o} - 1}\right)$$
which can be readily solved for r. Keep in mind the trig identity for tan(θ/2)

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