Energy for Orbits: 500km Above Earth's Surface

  • Thread starter Thread starter ice
  • Start date Start date
  • Tags Tags
    Energy Orbits
AI Thread Summary
To send a 1000kg object to a height of 500km above Earth's surface, approximately 4.5 billion Joules of energy is required just for lifting. However, to place the object into orbit at that height, additional energy is needed to achieve the necessary orbital velocity, which is significantly greater than the energy required for lifting alone. The discussion emphasizes that mechanical energy in orbit consists of both kinetic and potential energy, necessitating a calculation of their changes. The correct approach involves equating gravitational force with centripetal force to derive the orbital velocity. Understanding these energy requirements is crucial for accurately determining the energy needed for both lifting and orbital insertion.
ice
Messages
4
Reaction score
0
This question came up on a quiz and i got it wrong.
a)how much energy does it take to send a 1000kg object from the surface of the Earth to a height of 500km above the Earth's surface? b) how much energy does it take to put the same object into ORBIT at 500km above the surface of the earth.

For part A, i got -4.5*10^9 Joules, but i don't know if that's right.
part b is what i really don't understand, so any help here would be appreciated, thanks.
 
Physics news on Phys.org
In part A you simply calcuated the amount of work required to lift an object to a certain height. If you simply lifted it to that height and let it go it would simply fall back to Earth. In part B, the keyword is ORBIT which means that in addition to merely lifting the object you must impart to it sufficient velocity for it to remain at that height in order for it to ORBIT the Earth. You will find that the kinetic energy requirement of orbiting is substantially greater than the requirement for simply lifting the object.
 
but isn't Ek=1/2 Ep for all orbits? i tried using v^2 = sqrt(2GM/r), which didnt really work out.
 
I think for b) it is:

<br /> ME = \frac{1}{2}mv^2 - \frac{mMG}{r}<br />


<br /> ME = \frac{mMG}{2r} - \frac{mMG}{r}<br />

The Mechanical Energy = Kinetic Energy + Potential Energy
I substituted the orbital velocity for v which equals to:

<br /> v_{orb} = \sqrt{\frac{MG}{r}}<br />

You can derive this by putting the Gravitational Force to be equal to the centripetal force.

If there are any errors - feel free to comment.
 
The part that you're leaving out is that the payload starts off on the surface of the Earth with whatever potential energy it has there and (in the simplified case) NO kinetic energy. What you need to calculate is the CHANGE in both the payload's potential energy and kinetic energy.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top