The rocket launch and changing gravity

Click For Summary
SUMMARY

The discussion focuses on the physics of rocket launches, emphasizing the impact of Earth's rotational velocity on the required thrust for achieving orbit. A rocket launched eastward benefits from Earth's rotation, needing less energy to reach escape velocity, calculated as 10806.99 m/s after accounting for the rotational speed of 393.11 m/s. The energy required for a 1000 kg rocket to reach escape velocity is estimated at 5.84 x 10^10 J, considering gravitational changes and mass loss due to fuel consumption. Participants highlight the importance of averaging gravitational effects when calculating energy and weight at varying altitudes.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with gravitational physics and escape velocity calculations
  • Knowledge of energy equations, specifically E=1/2 mv^2
  • Basic concepts of mass and weight differentiation
NEXT STEPS
  • Research the effects of Earth's rotation on rocket launches
  • Study gravitational variations with altitude using g=GM/r^2
  • Explore the implications of fuel consumption on rocket mass and thrust
  • Learn about energy conservation in rocket propulsion systems
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in the mechanics of rocket launches and gravitational effects on propulsion.

marco12345a
Messages
13
Reaction score
0
When a rocket is launched, it starts not with velocity zero, but with the rotation velocity which the Earth gives it. Thus if a rocket is launched eastward, it requires a smaller boost (and if westward, a larger one) to achieve orbit. with a latitude of 31.1 degrees, cos(31degrees) 24902/24 = 393.11 ms/s

To escape velocity on Earth = (2G Mearth / r earth) = 11200ms-1
So the total velocity that needed for the rocket = 11200 - 393.11= 10806.99ms-1

I am trying to find the total energy that is needed for the rocket of 1000kg to escape velocity

i used the question
m = 1000 x 9.8 N

E= 1/2 m^2
= 5.84 X 10^10 J

my question is : as the further you are away from earth, there will be less gravity. So if i use that equation, would that be accurate assume there is no air friction and the mass of the rocket doesn't change due to the loss of fuel
 
Physics news on Phys.org
That would be close enough! But the problem is that the accuracy falls as u go on neglecting decreasing mass due to consumption of fuel.
 
should i use the equation of E= 1/2 m^2 ?
 
marco12345a said:
should i use the equation of E= 1/2 m^2 ?

If u r calculating energy, then of course it'll do. note that, E=1/2mv^2=GMm/r
 
but doesn't the changing gravity affect the m ?
 
marco12345a said:
but doesn't the changing gravity affect the m ?

Yes but if u really want to solve it u will have to take an average of that.
 
marco12345a said:
but doesn't the changing gravity affect the m ?

more than that...changing mass is a prob.
 
for example if a rocket is 1000 kg on Earth , what would the gravity be when it is at 300m
 
marco12345a said:
for example if a rocket is 1000 kg on Earth , what would the gravity be when it is at 300m
If u r dealing with mass, it would be 1000Kg itself. if u r dealing about wight, itll bcom
W=mg
=1000*g
g= gravitational acceleration at 300mts
u manipulate it...
g=w/m
=w/1000
ul have to know its weight there.
 
  • #10
how do you find the changing gravity ?
 
  • #11
  • #12
the gravity doesn't change preferentially at 300mts
observ: g=9.806 in 0mts
g=9.803at 1000mts

ok bye..
 
  • #13
okay thank you very much
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
12K