How Accurate Are Calculations for Satellite Orbit Energy and Rocket Altitudes?

  • Thread starter Thread starter six789
  • Start date Start date
  • Tags Tags
    Energy Satellites
AI Thread Summary
The discussion focuses on calculating the change in gravitational potential energy for a satellite moving to a 2500 km orbit. The initial calculations yielded a change of approximately -1.27 x 10^11 J, while the book states the answer should be 1.1 x 10^11 J. The error was identified in the gravitational potential energy formula used, suggesting the need for consistency in the calculations. Additionally, a separate problem regarding a rocket's launch velocity of 3.4 km/s raised questions about determining its maximum height above Earth's surface, with the correct answer being 650 km. The conversation emphasizes the importance of using the correct formulas and values in orbital mechanics calculations.
six789
Messages
127
Reaction score
0
this is the problem...
What is the change in gravitational potential energy of a 6200kg satellite that lifts off from Earth's surface into a circular orbit of altitude2500km?

i did this...
r = rE +h
=6.38 x10^6m + 2.5 x10^6m
r=8.88 x10^6m

Eg= -GMm
r
=-6.673x10-11N m2/kg2 (5.98 x10^24kg)(6200kg)
8.88 x10^6m
Eg=-2.786127793 x10^11J

Eg=mgh
=6200kg(9.82m/s2)(-2.5 x10^6m)
Eg=-1.52055 x10^11J

Change in Gravitational Potential Energy
∆ Eg = Eg - Eg
=-2.786127793 x10^11J – (-1.52055 x10^11J)
∆ Eg =-1.265577793 x10^11J

but the answer on my book is 1.1x10^11J, can anyone check this, coz I am not sure where is the mistake.. thanks for the help
 
Last edited:
Physics news on Phys.org
Your problem is here:

Eg=mgh
=6200kg(9.82m/s2)(-2.5 x10^6m)
Eg=-1.52055 x10^11J

You should have just the used the same form used in the prior Eg formula

E_g = \frac{GMm}{R}

with a different value for R
 
satellites

thanks for the help... i get the answer now...
 
another problem is... a rocket is launched vertically from Earth's surface with a velocity of3.4km/s. How high it go a)from Earth's centre and b) from Earth's surface.
a)r = 2GM
V^2
= 2(6.673x10^-11N m^2/kg^2) (5.98 x10^24kg)
(3400m/s)^2
r=7.0x107m

i don't know waht to do with b), i am thinking to add the raduis of the earth, but my answer is wrong... the answer should be 650km.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top