How High Will a Projectile Go if Fired at 8.2 km/s from Earth?

In summary, the conversation discusses solving for the final distance above Earth's surface that a projectile will travel when fired vertically with an initial speed of 8.2 km/s. The formula used is conservation of energy, where the mass of the projectile is not needed and cancels out. The final distance is solved for using the final velocity at the highest point, the initial velocity, and the initial and final distances from Earth's center. The final answer is in meters and should be subtracted by Earth's radius to get the distance above the surface.
  • #1
CinderBlockFist
86
0
A projectile is fired vertically from Earth's surface with an initial speed of 8.2 km/s. Neglecting air drag, how far above the surface of Earth will it go?


I am completely stuck. What formula should i use, this was in the Gravitational Potential Energy /escape speed section
 
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  • #2
use conservation of energy here...
warning: the potential energy is not mgh in this case
 
  • #3
vince, how do i use conservation of energy if i am not given the mass of the projectile? thanks
 
  • #4
you don't need the mass, since the KE and the PE is directly proportional to the mass, the mass will cancel out anyway... just put the mass = m in the formulas... the m will not appear in your final answer
 
  • #5
but then if i just put m, i have too many unknowns, i don't know the final velocity, only the initial. so i am stuck
 
  • #6
and what variable should i be solving for if i want to know the final distance from earth?
 
  • #7
Write down the equations first, Cinder, then complain after they don't work.

And if you want to know the final distance from Earth, you should probably be solving for the final distance from Earth.

--J
 
  • #8
I have equations, this is what I am using, no need to be so sarcastic, geeze. Sorry if i don't understand it so well.

K(final)+ U(final) = K(initial)+ U(initial)


1/2mvfinal^2 - GMm/R = 1/2 Mvinitial^2 - GMm/R

what i meant was, which one of these variables give the distance from earth? whiich should i solve for?
 
  • #9
yeah, you have some unknown...
mass of the projectile, which I have already told you it doesn't matter
initial velocity, isn't that given?
final velocity, at the highest point, do you still have velocity?
R initial... your projectile fire from the surface of earth... so R is...
R final... this is your only unknown... and also your answer... solve for it

and the M... mass of earth... is that an unknown?
 
Last edited:
  • #10
ah crap, i was thinking i didnt have v final, but i guess i had to assume it stops so it is zero. thanks vince, brb leme do my calculations.
 
  • #11
damn how come when i try to isolate R final, the units end up w/ m^2/s^2 * meters, so i end up w/ m/s^2 when i should just get meters for the final radius, correct?


when i isolate Rfinal i get -GMm/(1/2mvinitial^2-Gmm/Riniitial) theni cancel out the masses like u said vince, iand i get

R final = -GM/ (1/2 Vinitial^2- GM/R initial) and the units arent meters, its m/s^2
 
  • #12
apparently, the unit is fine...you have GM/(***+GM/R) the GM cancels out and leave the R alone.. how could you get m/s^2?
and your equation is fine too...
 
  • #13
ok, leme just carry out the number calculations, and see what i get. brb
 
  • #14
in the denominator, I have the kinetic energy still , so my units arrent matchin up to meters on the left hand side.



R final = -GM
------------
(1/2 Vinitial^2 - GM/ Rinitial)
 
  • #15
I see the GM's cancel out in the numerator and denominator. But then i have the 1/2 V initial^2 - 1/ R initial. So that is m/s^2 - meters, i don't get it.
 
  • #16
what is the unit of GM/v^2... check your data book what is the unit of G
 
  • #17
crap nevermind my algebra was bad! forgot to cancel out the units w/ the first term (kinetic units) in the denomitaor as well, ok brb sorry.
 
  • #18
Vince, since it asks how far about the surface of the Earth will it go, do i subtract the radius of the Earth from my R final? Cause isn't R final the distance from the middle of the Earth to the middle of the projectile.
 
  • #19
Vince, thank you man i got it! I really appreciate your help.
 

1. What is a projectile?

A projectile is any object that is launched into the air and follows a curved path under the influence of gravity.

2. How does a projectile fired from Earth's surface behave?

A projectile fired from Earth's surface will follow a curved path due to the pull of gravity, eventually landing on the ground. The shape of the path is called a parabola.

3. What factors affect the trajectory of a projectile fired from Earth's surface?

The trajectory of a projectile is affected by the initial velocity of the object, the angle at which it is fired, and the force of gravity.

4. Can a projectile fired from Earth's surface achieve orbit?

No, a projectile fired from Earth's surface cannot achieve orbit because it does not have enough horizontal velocity to counteract the force of gravity and maintain a circular path around the Earth.

5. How does air resistance affect a projectile fired from Earth's surface?

Air resistance can affect the trajectory of a projectile by slowing it down and causing it to fall to the ground more quickly. This can be minimized by reducing the surface area of the projectile and increasing its initial velocity.

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