# Launch - universal gravitation

1. Apr 25, 2010

### mybrohshi5

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

A small package is fired off Earth's surface with an unknown launch speed, but with a known launch angle of 51.0 as measured from the local horizontal. It reaches a maximum height above the surface of 6380.0 km, a value equal to Earth's radius itself. What is its speed when it reaches this height? Ignore any effects that might come from Earth's rotation or from air resistance.

What is the speed of the package when it reaches its maximum height?
What was the launch speed of the package?

2. Relevant equations

Ki + Ui = Kf + Uf

U = -Gmm/r

3. The attempt at a solution

For the first question: What is the speed of the package when it reaches its maximum height?

0 = Kf + Uf

0 = 1/2mv2 + -GMem / r

GMe / r = 1/2v2

v = 11181.96 m/s

Does this seem right? i fee like its wrong cause a hint the question gives is: (Hint: You will need to use two conservation equations and solve two equations in two unknowns.)

Thanks for any help

2. Apr 25, 2010

### Andrew Mason

What is the expression for kinetic energy in terms of vertical and horizontal components of velocity? Does the kinetic energy relating to the horizontal velocity contribute to the height of the trajectory?

That should enable you to determine the kinetic energy at launch from the vertical component of the launch velocity from the maximum height of the trajectory? That will give you the vertical speed. Since you know the angle of launch, you can then determine horizontal speed (is there any vertical speed at maximum height?).

AM

3. Apr 26, 2010

### mybrohshi5

This is what i did new to find: What is the speed of the package when it reaches its maximum height?

x=theta

1/2m(vsinx)2 = mgh

1/2v2sinx2 = gh

solve for v to get 455.025 km/s

then to find the x component of the velocity

vx = 455.025(cos51)

vx = 286.35 km/s

4. Apr 26, 2010

### Andrew Mason

You have the right approach. The initial vertical kinetic energy has to equal the change in gravitational potential energy at maximum height (when vertical speed = 0). The only problem with your method is in calculating that change in gravitational potential energy.

For small changes in height, the force of gravity does not change very much. So the change in gravititational PE is approximately mgh.

But for large changes in height (such as this), the change in force of gravity is material. You have to apply Newton's law of universal gravitation to find the change in potential energy.

AM

5. Apr 26, 2010

### mybrohshi5

I used -GMm/r now for my gravitational potential and this is what i came up with

x=theta

Me = mass of the earth = 5.97x10^24

1/2m(vsinx)2 = GMm/r

1/2v2sinx2 = GMe/r

i thought r should be 6380km + 6380km since radius of the earth is 6380 and it goes to a height of 6380 so the total r should be 12760km total?

solve for v to get 321467 km/s

then to find the x component of the velocity

vx = 321467(cos51)

vx = 202306 km/s

Am i still off?

Thanks for the help

6. Apr 26, 2010

### D H

Staff Emeritus
Please try again, from the start.

You don't need to know about the different components of the velocity to determine the kinetic energy. Kinetic energy is $mv^2/2$ regardless of the direction of the velocity vector.

You have not used the hint yet. You have only used conservation of energy. What other conservation laws apply here? (Hint: You will need to worry about the direction of the velocity vector for this other law.)

7. Apr 26, 2010

### mybrohshi5

I cant really think of the other law that applies. Would it maybe be the law of conservation of momentum?

8. Apr 26, 2010

### Andrew Mason

Your calculation of change in potential is not correct.

$$\Delta PE = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R}$$

BTW, I am not sure about DH's comments. It appears to me that you are using the launch angle correctly.

AM

Last edited: Apr 26, 2010
9. Apr 26, 2010

### mybrohshi5

Ok i will give it another shot and see if i can get it using the correct potential now :)

10. Apr 26, 2010

### D H

Staff Emeritus
He's not. Let's denote the initial speed (the magnitude of the initial velocity) as vi and the final speed as vf. Denoting $\hat x$ as the local horizontal and $\hat y$ as local vertical, the initial velocity is

$$\mathbf v_i = v_i (\cos 51^{\circ} \hat x + \sin 51^{\circ} \hat y)$$

The kinetic energy is $1/2 m \mathbf v_i \cdot \mathbf v_i =1/2 m v_i^2(\cos^2 51^{\circ} +\sin^2 51^{\circ}) = 1/2 m v_i^2\$. The launch angle does not come into play in the calculation of the initial kinetic energy (or any kinetic energy).

Rather than working in cartesian coordinates ($\hat x$ and $\hat y$) it will make a bit more sense to work in polar coordinates ($\hat r$ and $\hat \theta$) -- particularly at the desired maximum altitude point.

11. Apr 26, 2010

### cjpgconman

So I'm having trouble on this problem as well:
I have trouble visualizing it so is this the full work energy equation:
(1/2)m*vi^2+((GMm)/R)=(1/2)*m*vf^2-(GMm)/R
I want to say this is right but I don't know, and if it is what is the other conservation equation? I think it would be momentum but I really don't know

12. Apr 26, 2010

### Andrew Mason

But it is only the radial component of velocity that matters here: $\Delta PE = F\cdot dr = Fsin\theta dr$

So:

$$\frac{1}{2}mv_{yi}^2 = \Delta PE = \frac{GMm}{2R}$$

The tangential component does not change so when vy is 0 (at maximum height) the kinetic energy is just:

$$KE = \frac{1}{2}mv_x^2 = \frac{1}{2}m(v_0\cos\theta)^2$$

AM

13. Apr 26, 2010

### D H

Staff Emeritus
Linear momentum? Think about it for a bit in terms of a circular orbit, particularly the momentum at opposite points on the orbit. Another example: An object is launched from the Earth with an initial velocity exactly equal to escape velocity. What happens to the velocity as the object gets further and further from the Earth.

What other conservation laws are there?

Last edited: Apr 26, 2010
14. Apr 26, 2010

### D H

Staff Emeritus
That is not correct, Andrew. I sent you a PM.

15. Apr 26, 2010

### D H

Staff Emeritus
You have the correct kinetic energy but the potential energy is wrong on both sides of the equality. Why would the potential change signs, and why are both potentials divided by the radius of the Earth?

16. Apr 26, 2010

### cjpgconman

oh dangit this is what it should be (I think)
(1/2)m*vi^2-((GMm)/R)=(1/2)*m*vf^2-(GMm)/2R

and for the other conservation law would it be orbital momentum? our teacher hasn't really gone over it but we have homework on it, so I get the idea I just have no idea how to use that kind of momentum
sorry if I'm just not getting it

17. Apr 26, 2010

### D H

Staff Emeritus
Right result, how did you get there?

What are some other conservation laws? (Use Google if you must.) Which ones apply here?

18. Apr 26, 2010

### mybrohshi5

I am working on it right now and i think i got it. If i get it right i will let you know.

19. Apr 26, 2010

### D H

Staff Emeritus
If you do get it soon (and I mean soon! it's 11:30 PM here; way past time to go to bed) I will check your results. Otherwise you may have to wait. I'll post the solution for our homework helpers just in case someone else is up late.

20. Apr 26, 2010

### mybrohshi5

Ok i came up with this and i am postive about it all except for one little thing.

Use conservation of energy like you had.

1/2mvi2 + -GMm/r = 1/2mvf2 + -GMm/2r

then use the conservation of angular momentum

Li = Lf

mvi(cos51)r = mvf(2r)

The part i am unsure about is the angular momentum equation

i dont get why you would use vi(cos51) for the initial and then only vf for the final. maybe DH can explain this part :)

but anyways you can solve for vf from the angular momentum equation and then plug that into the energy equation and then solve for vi.

since vi is then found you can plug that into vf = (vi/2)(cos51)

that was found when you solved for the vf from the angular momentum equation.