Stuck on projectile motion problem using gravitational force

[giantf]

Homework Statement

A projectile is fired vertically from the Earth's surface with an initial speed of 11.4 km/s. Neglecting air drag, how far (in meters) above the surface of the Earth will it go?

Homework Equations

(1/2)mv^2
-GmM/R

The Attempt at a Solution

KE_i+PE_i = KE_f+PE_f
(1/2)mv^2 - GmM/R = -GmM/(R+h)
R+h = -(GM)/((1/2)v^2-GM/R)
h = -(GM)/((1/2)v^2-GM/R) - R
and plugging everything in, I got -1.744x10^8
So I tried using kinetic and potential energy to solve for h, but keep getting a negative number. Am I doing this wrong?

 

[gneill]

Have you learned about the implications of the sign of the total mechanical energy value?

Try this in your conservation of energy equation: Rather than assuming that the final KE goes to zero, suppose that the value of h grows large enough that the PE tends to zero. What value does that yield for the final KE?

 

[giantf]

Have you learned about the implications of the sign of the total mechanical energy value?

Try this in your conservation of energy equation: Rather than assuming that the final KE goes to zero, suppose that the value of h grows large enough that the PE tends to zero. What value does that yield for the final KE?

I don't think we've learned about that yet, and how would I leave final KE in? I don't think I'm supposed to find the final velocity

 

[gneill]

I don't think we've learned about that yet, and how would I leave final KE in? I don't think I'm supposed to find the final velocity

The idea is to investigate another possible scenario.

When you set the final KE to zero you are making the assumption that the object will eventually slow to zero before falling back. What if the velocity never never reaches zero? What will be the maximum height then? So instead of assuming that the KE goes to zero, assume that the PE goes to zero instead. How will your equation look?

 

[giantf]

The idea is to investigate another possible scenario.

When you set the final KE to zero you are making the assumption that the object will eventually slow to zero before falling back. What if the velocity never never reaches zero? What will be the maximum height then? So instead of assuming that the KE goes to zero, assume that the PE goes to zero instead. How will your equation look?

would have
(1/2)(11400^2)-((6.67*10^-11)(5.98*10^24))/(6.371*10^6)=(1/2)v_f^2
which would make v_f = 2178.762
but I don't see where I can get the h variable from if i set potential final energy to zero

 

[gneill]

would have
(1/2)(11400^2)-((6.67*10^-11)(5.98*10^24))/(6.371*10^6)=(1/2)v_f^2
which would make v_f = 2178.762

Good.

but I don't see where I can get the h variable from if i set potential final energy to zero

Well, if the velocity never, ever, goes to zero, what can you say about h?

 

[giantf]

Good.

Well, if the velocity never, ever, goes to zero, what can you say about h?

It never stops, infinity? but the question is looking for a specific number

 

[gneill]

It never stops, infinity? but the question is looking for a specific number

Right, it never stops. There's no specific (finite) value for h. h tends to infinity. That's the best answer you can give.

Some things to investigate:
Look up the concept of escape velocity. You should be able to find or compute a value for the escape velocity from the surface of the Earth. Compare it to your given value for the initial velocity.

Compute the specific mechanical energy for the object: That's your energy formula without the mass of the object in question so that you end up with the units joules per kg: ##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_o}##. There are three cases:

1) If ξ is negative then the object is "bound": it will never escape. Its speed will eventually reach zero at some finite distance then fall back to Earth.

2) If ξ is precisely zero it has exactly escape velocity. Its speed will never reach exactly zero, but it will approach zero in the limit. Again, it will never fall back and the "final" distance is infinity.

3) If ξ is a positive value then again the object will escape, and moreover, its velocity will approach some value greater than zero as its distance goes to infinity (this "final" velocity is sometimes called the excess velocity).

 

[giantf]

Right, it never stops. There's no specific (finite) value for h. h tends to infinity. That's the best answer you can give.

Some things to investigate:
Look up the concept of escape velocity. You should be able to find or compute a value for the escape velocity from the surface of the Earth. Compare it to your given value for the initial velocity.

Compute the specific mechanical energy for the object: That's your energy formula without the mass of the object in question so that you end up with the units joules per kg: ##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_o}##. There are three cases:

1) If ξ is negative then the object is "bound": it will never escape. Its speed will eventually reach zero at some finite distance then fall back to Earth.

2) If ξ is precisely zero it has exactly escape velocity. Its speed will never reach exactly zero, but it will approach zero in the limit. Again, it will never fall back and the "final" distance is infinity.

3) If ξ is a positive value then again the object will escape, and moreover, its velocity will approach some value greater than zero as its distance goes to infinity (this "final" velocity is sometimes called the excess velocity).

so i did sqrt(2GM/R) and plugged into v₁² and got

ξ = 63816956.64

 

[gneill]

so i did sqrt(2GM/R) and plugged into v₁² and got

ξ = 63816956.64

You need to include the potential energy, too.

 

[giantf]

You need to include the potential energy, too.

You need to include the potential energy, too.

got 99716437393501.805

 

[gneill]

got 99716437393501.805

Okay, first, it's a bit silly to quote so many digits. Use exponential notation when the numbers go over 10³.
Second, what calculation did you do to include the PE? If you started with escape velocity I would expect your ξ to end up at or very close to zero.

 

[giantf]

Okay, first, it's a bit silly to quote so many digits. Use exponential notation when the numbers go over 10³.
Second, what calculation did you do to include the PE? If you started with escape velocity I would expect your ξ to end up at or very close to zero.

Okay, first, it's a bit silly to quote so many digits. Use exponential notation when the numbers go over 10³.
Second, what calculation did you do to include the PE? If you started with escape velocity I would expect your ξ to end up at or very close to zero.

I used the ξ equation and plugged the escape velocity in v_f²

 

[gneill]

I used the ξ equation and plugged the escape velocity in v_f²

But there's no ##v_f## in:

##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_i}##

 

[giantf]

But there's no ##v_f## in:

##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_i}##

i meant V initial

 

[gneill]

i meant V initial

Something is amiss. Can you write out the values you're using for each of the variables?

 

[giantf]

Something is amiss. Can you write out the values you're using for each of the variables?

ξ=(sqrt(2(6.67*10^-11 * 5.98*10²⁴))^2/2 - (6.67*10^-11 * 5.98*10²⁴)/6.371*10⁶

 

[gneill]

ξ=(sqrt(2(6.67*10^-11 * 5.98*10²⁴))^2/2 - (6.67*10^-11 * 5.98*10²⁴)/6.371*10⁶

You forgot to divide by R in the first term. The escape velocity is given by

##v_{esc} = \sqrt{\frac{2 G M}{R}}##

 

[giantf]

You forgot to divide by R in the first term. The escape velocity is given by

##v_{esc} = \sqrt{\frac{2 G M}{R}}##

i got -6.26*10^7

 

[giantf]

i got -6.26*10^7

actually nvm forgot to square it, i got 0

 

[giantf]

actually nvm forgot to square it, i got 0

but it's not accepting it as my answer, it's looking for a height

 

[gneill]

actually nvm forgot to square it, i got 0

If you'd done it symbolically you could have avoided a lot of calculator button pushing:

##ξ = \frac{1}{2}\left(\sqrt{\frac{2 G M}{R}}\right)^2 - \frac{G M}{R}##

##ξ = \frac{1}{2}\frac{2 G M}{R} - \frac{G M}{R}##

##ξ = \frac{G M}{R} - \frac{G M}{R}##

##ξ = 0##

 

[gneill]

but it's not accepting it as my answer, it's looking for a height

Of course. Your launch speed is greater than the escape velocity. So there is no maximum height: it's infinite. I thought you were simply looking to check the assertion that the mechanical energy is zero when the launch speed is the escape velocity. I didn't realize that you were still looking for some finite value for h. There isn't one for the given initial speed.

If the system is expecting a finite value then either the question is flawed (they gave you an inappropriate initial velocity) or there must be some way to enter "infinity" as a response. Can you verify that the initial velocity is 11.4 km/s?

 

[giantf]

Of course. Your launch speed is greater than the escape velocity. So there is no maximum height: it's infinite. I thought you were simply looking to check the assertion that the mechanical energy is zero when the launch speed is the escape velocity. I didn't realize that you were still looking for some finite value for h. There isn't one for the given initial speed.

If the system is expecting a finite value then either the question is flawed (they gave you an inappropriate initial velocity) or there must be some way to enter "infinity" as a response. Can you verify that the initial velocity is 11.4 km/s?

The hint for the problem was " Kinetic energy is converted into gravitational potential energy. Don't forget to subtract the radius of the Earth from your final distance to determine the final distance above the surface of the earth."
I thought I was doing it right, as my original answer did subtract the radius

 

[gneill]

View attachment 197615
The hint for the problem was " Kinetic energy is converted into gravitational potential energy. Don't forget to subtract the radius of the Earth from your final distance to determine the final distance above the surface of the earth."
I thought I was doing it right, as my original answer did subtract the radius

Your approach was fine. The problem is that question is flawed. You should report it to your instructor.

 

[giantf]

Your approach was fine. The problem is that question is flawed. You should report it to your instructor.

thanks for everything, it was a confusing one