Launch - universal gravitation

[mybrohshi5]

Homework Statement

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?

Homework Equations

Ki + Ui = Kf + Uf

U = -Gmm/r

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 + -GM_em / r

GM_e / 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

 

[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

 

[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

v_x = 455.025(cos51)

vx = 286.35 km/s

Am i still going about this wrong?

 

[Andrew Mason]

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

v_x = 455.025(cos51)

vx = 286.35 km/s

Am i still going about this wrong?

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

 

[mybrohshi5]

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

x=theta

M_e = mass of the Earth = 5.97x10^24

1/2m(vsinx)² = GMm/r

1/2v²sinx² = GM_e/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

 

[D H]

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 [itex]mv^2/2[/itex] 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.)

 

[mybrohshi5]

I can't really think of the other law that applies. Would it maybe be the law of conservation of momentum?

 

[Andrew Mason]

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

x=theta

M_e = mass of the Earth = 5.97x10^24

1/2m(vsinx)² = GMm/r

1/2v²sinx² = GM_e/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

Your calculation of change in potential is not correct.

[tex]\Delta PE = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R}[/tex]

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

AM

 

[mybrohshi5]

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

 

[D H]

He's not. Let's denote the initial speed (the magnitude of the initial velocity) as v_i and the final speed as v_f. Denoting [itex]\hat x[/itex] as the local horizontal and [itex]\hat y[/itex] as local vertical, the initial velocity is

[tex]\mathbf v_i = v_i (\cos 51^{\circ} \hat x + \sin 51^{\circ} \hat y)[/tex]

The kinetic energy is [itex]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\ [/itex]. 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 ([itex]\hat x[/itex] and [itex]\hat y[/itex]) it will make a bit more sense to work in polar coordinates ([itex]\hat r[/itex] and [itex]\hat \theta[/itex]) -- particularly at the desired maximum altitude point.

 

[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

 

[Andrew Mason]

He's not. Let's denote the initial speed (the magnitude of the initial velocity) as v_i and the final speed as v_f. Denoting [itex]\hat x[/itex] as the local horizontal and [itex]\hat y[/itex] as local vertical, the initial velocity is

[tex]\mathbf v_i = v_i (\cos 51^{\circ} \hat x + \sin 51^{\circ} \hat y)[/tex]

The kinetic energy is [itex]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\ [/itex]. 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 ([itex]\hat x[/itex] and [itex]\hat y[/itex]) it will make a bit more sense to work in polar coordinates ([itex]\hat r[/itex] and [itex]\hat \theta[/itex]) -- particularly at the desired maximum altitude point.

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

So:

[tex]\frac{1}{2}mv_{yi}^2 = \Delta PE = \frac{GMm}{2R}[/tex]

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

[tex]KE = \frac{1}{2}mv_x^2 = \frac{1}{2}m(v_0\cos\theta)^2[/tex]

AM

 

[D H]

what is the other conservation equation? I think it would be momentum but I really don't know

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?

 

[D H]

But it is only the radial component of velocity that matters here

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

 

[D H]

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

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?

 

[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

 

[D H]

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

Right result, how did you get there?

and for the other conservation law would it be orbital momentum?

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

 

[mybrohshi5]

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

 

[D H]

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.

 

[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.

1/2mv_i² + -GMm/r = 1/2mv_f² + -GMm/2r

then use the conservation of angular momentum

L_i = L_f

mv_i(cos51)r = mv_f(2r)

The part i am unsure about is the angular momentum equation

i don't get why you would use v_i(cos51) for the initial and then only v_f for the final. maybe DH can explain this part :)

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

since v_i is then found you can plug that into v_f = (v_i/2)(cos51)

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

 

[D H]

Angular momentum is the vector cross product [itex]\mathbf r \times \mathbf v[/itex]. What are the angles between the radial and velocity vectors initially and at the maximum altitude?

 

[mybrohshi5]

I want to say something but to be honest i have never really understood cross product so i really am not sure what the angles are between the radial velocity vectors initially and at the maximum altitude :(

I don't want to keep you up since its so late there (only 9:30pm here) so if you want to explain this tomorrow that would be more than fine :) thank you for all the help

 

[D H]

It is far too late for me here. Perhaps someone else can fill in. If not, I'll try to write something tomorrow AM.

 

[mybrohshi5]

No problem. Thanks again for the help. I appreciate it.

 

[cjpgconman]

Ok so I didn't get it when I plugged it in with your equations (but there was a lot of scribbling so it was probably calc error)
but I got it using these equations

Vi*sin(90-theta)=2Vf
and
Vf^2=Vi^2-GM/R

two equations and two unknowns just make sure your units are right
Thanks for all the help guys

 

[D H]

The magnitude of the cross product [itex]\mathbf r \times \mathbf v[/itex] will be rv sinθ, where r and v are the magnitudes of the position and velocity vectors and θ is the angle (between 0 and 180º) between these two vectors. At launch, [itex]\mathbf r[/itex] is local vertical. You are given the information needed to determine θ at launch. There is only one possible value for θ at the point where r attains its maximal value. What is that angle?

 

[ehild]

The magnitude of the cross product is rv*sin(theta)

ehild

 

[D H]

Thanks, corrected. That was a typo in converting from latex to inline.

 

[Andrew Mason]

DH is correct. I was over simplifying the problem. This makes it a much more interesting problem. Thanks DH.

Angular momentum is conserved, which means that the tangential speed decreases by half at maximum height (2R) from the tangential speed at launch.

AM

 

[mybrohshi5]

The magnitude of the cross product [itex]\mathbf r \times \mathbf v[/itex] will be rv sinθ, where r and v are the magnitudes of the position and velocity vectors and θ is the angle (between 0 and 180º) between these two vectors. At launch, [itex]\mathbf r[/itex] is local vertical. You are given the information needed to determine θ at launch. There is only one possible value for θ at the point where r attains its maximal value. What is that angle?

Since the given angle is 51 degrees would the angle between the r and v be 39?

I am trying to picture what you are saying and i looked in my book about vector product but i am still having a hard time grasping it i guess.

 

[D H]

It often helps to draw pictures. But yes, the angle between the radial and velocity vectors would be 39 degrees at launch. What is the angle going to be at the maximum altitude?

Think about it, sketch some pictures. It might help to bring the picture a bit closer to Earth. For example, imagine a baseball thrown from center field to home. At the top of the baseball's arc, what is the angle between the ball's velocity vector and local vertical?

 

[mybrohshi5]

Isnt it 90 degrees because at the top of the arc (or maximum altitute) there is only velocity in the horizontal direction and none in the vertical?

 

[D H]

Correct.

What's next? Can you finish this problem now?

 

[mybrohshi5]

Yes that helps. I am going to rework the problem this weekend when i am studying for my test and if i run into any problems i will post here again.

Thank you for all the help.