Understanding Forces and Energy in Space Exploration

  • Thread starter Thread starter m00c0w
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating forces and energy in space exploration scenarios. A participant calculated the force on a rocket using F=ma but was marked wrong, prompting questions about gravitational effects during acceleration. Another participant calculated the kinetic energy of an astronaut jumping on the moon and expressed uncertainty about the relevance of the drop distance in the question. The conversation also explores the differences in jumping safely on the moon versus Earth, highlighting the impact of lower gravitational field strength on weight and energy. Lastly, the benefits of underwater training for astronauts are discussed, emphasizing the simulation of weightlessness and multidimensional movement.
m00c0w
Messages
17
Reaction score
0
Hi I'm doing a practice test paper in preparation for my exam. I've calculated the acceleration of a rocket to be 5m/s². It now says that the average mass of the rocket is 1000kg and asks to calculate the force on the rocket. I assumed that the formula to use would be F=ma so I calculated the force to be 5000N from 1000 x 5 although this was marked wrong. Could anyone please explain where I went wrong?

Then there is a question on an astronaut. He jumps off a rock on the moon and drops 1.5m. His speed is 2.2m/s as he hits the ground. Calculate the kinetic energy of the astronaut as he hits the ground. I used the formula K.E. = 1/2 x mass x speed² to get 1/2 x 150 x 2.2² = 363J. I was wondering if anyone could confirm whether that is right as I have a suspicion it isn't considering they bothered to mention the distance he drops in the question.

Then I have to explain why astronauts can jump down from a height of 1.5m safely on the moon but not safely on earth. I have to make reference to weight, acceleration and energy. So far I've said that on the moon the astronauts weight will be less as the gravitational field strength will be less. Also the astronaut will have less gravitational potential energy as the gravitational field strength is less. Is there a scientific way that I can tie in acceleration in relation to the gravitational field strength?

Finally there is a question asking why it is a good idea for astronauts to do training under water for their work on the moon as opposed to just on the ground on earth. I have forgot the technical words but I'm assuming it is to do with the drag in the water making them move slower :confused:

Any help would be greatly appreciated. Thanks in advance.
 
Physics news on Phys.org
No. 1 - are you told how long the rocket travels for? And what about the gravitational force on the rocket?

No. 3 - the acceleration and g and equal I think.

No. 4 - underwater is weightlessness simulation, no effects of gravity, can move in 3 dimensions etc.

-NewScientist
 
Thanks for the reply. Referring to No. 1 I am told that the rock travels in space. It accelerates from a velocity of 100m/s to 700m/s. This takes 120 seconds.
 
Well, it may be a trick question does it ask for the force after acceleration? For if it does the answer is 0 as the rocket will be at a steady speed and so will have acceleration of 0 and therefore the forces are balanced or 0. If they are balanced the net force is 0

-NewScientist
 
EDIT: nevermind :P
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top