Launched Rocket (Energy problem)

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In summary, the conversation discusses the calculation of a rocket's height above Earth's center when launched with a velocity of 3.4 km/s. Using the equations for kinetic and potential energy, the negative sign in the resulting equation can be ignored as it cancels out in the final calculation. To find the rocket's distance from Earth's surface, the radius of the Earth can be subtracted from the calculated value.
  • #1
I Like Pi
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Homework Statement



15. (4) A rocket is launched vertically from Earth's surface with a velocity of 3.4 km/s. How high does it go from Earth's centre?


Homework Equations



Ek = 1/2mv^2
Eg = -GMm/r

The Attempt at a Solution



1/2mv^2 = -GMm/r
1/2(3400)^2 = 6.673*10^-11*5.978*10^24/r
r = -6.9*10^11 m

How do i get rid of the negative?

Thanks for your time
 
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  • #2
Remember Eg is the energy needed to move the mass to infinity, so the change in Eg is the change in ke.

The change in PE is positive, and the change in KE is 0.5mv2

So you will have 0.5mv2= GMm/r
 
  • #3
rock.freak667 said:
Remember Eg is the energy needed to move the mass to infinity, so the change in Eg is the change in ke.

The change in PE is positive, and the change in KE is 0.5mv2

So you will have 0.5mv2= GMm/r

Thanks! so I just ignore the negative sign?
 
  • #4
I Like Pi said:
Thanks! so I just ignore the negative sign?

Well you don't really ignore is as it cancels out. But in essence, yes you can ignore it.
 
  • #5
rock.freak667 said:
Well you don't really ignore is as it cancels out. But in essence, yes you can ignore it.

How does it cancel out?

thanks for your help!
 
  • #6
I think it is a little more complicated. Notice that -GMm/r is zero at r = infinity (as someone already wrote) and that it gets larger as r increases. So, quite a large negative value at the surface. A smaller negative value a few km up. It really only has meaning when you take the difference for two different radii. You'll need something like this:
0.5mv² = -GMm/r - (-GMm/r)
where one of the r's is the Earth's radius and the other the radius you are looking for.
 
  • #7
Delphi51 said:
I think it is a little more complicated. Notice that -GMm/r is zero at r = infinity (as someone already wrote) and that it gets larger as r increases. So, quite a large negative value at the surface. A smaller negative value a few km up. It really only has meaning when you take the difference for two different radii. You'll need something like this:
0.5mv² = -GMm/r - (-GMm/r)
where one of the r's is the Earth's radius and the other the radius you are looking for.

Thank you :smile:! Makes sense. Now if i want to find the rocket's distance from EARTH'S SURFACE, would I just subtract the radius of the Earth from the answer i get above?

Thanks again!
 
Last edited:
  • #8
Right on the subtraction.
Most welcome and good luck on the next one.
 

FAQ: Launched Rocket (Energy problem)

How is energy used in a launched rocket?

Energy is used in multiple ways during a rocket launch. First, chemical energy from the rocket's fuel is converted into thermal energy, which creates hot gases that propel the rocket upwards. Then, electrical energy is used to power the various systems and instruments on board the rocket, such as the guidance and navigation systems. Finally, potential energy is converted into kinetic energy as the rocket gains speed and altitude.

What are the main energy sources for a launched rocket?

The main energy sources for a launched rocket are chemical energy and electrical energy. Chemical energy is used to power the rocket engines, while electrical energy is used for various systems and instruments on board.

How does the rocket overcome the energy problem of Earth's gravity?

The rocket overcomes the energy problem of Earth's gravity by using a combination of high thrust and high speed. The rocket engines provide a high amount of thrust, which counteracts the force of gravity pulling the rocket down. The rocket also gains speed as it travels upwards, which helps to overcome the pull of Earth's gravity.

What happens to the energy of a launched rocket after it reaches space?

After a launched rocket reaches space, the energy is still present but is no longer being used to overcome the force of Earth's gravity. Instead, the energy is used to maintain the rocket's speed and trajectory in space. Some of the energy may also be used to power onboard systems and instruments.

How does the energy problem affect the design of a launched rocket?

The energy problem plays a crucial role in the design of a launched rocket. Engineers must carefully consider the amount of energy needed to overcome Earth's gravity and reach the desired altitude and speed. This affects the type and amount of fuel used, the size and power of the rocket engines, and the overall design of the rocket and its systems. Additionally, the energy problem must be continuously monitored and managed throughout the rocket's launch and flight to ensure a successful mission.

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