How much mass will be converted to energy to accelerate the spaceship

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Homework Help Overview

The discussion revolves around a problem involving relativistic physics, specifically the mass-energy equivalence and the kinetic energy of a spaceship traveling at a significant fraction of the speed of light. The original poster presents a scenario where a spaceship with a total mass of 1.8×105 kg aims to travel to a star 30 light-years away at a speed of 0.70c, seeking to determine how much mass is converted to energy for acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for kinetic energy and the implications of mass change during acceleration. Some question the assumption of constant mass, while others explore the relationship between energy and mass conversion in the context of relativistic speeds.

Discussion Status

The conversation is ongoing, with several participants providing insights and corrections regarding the calculations and assumptions made by the original poster. There is a recognition of the complexity involved in the problem, particularly concerning the changing mass of the spaceship and the implications for energy calculations.

Contextual Notes

Participants note the lack of specific details such as exhaust velocity or specific impulse, leading to assumptions about energy conservation and mass loss. The original poster's interpretation of the problem is also discussed, highlighting potential ambiguities in the question's wording.

AyooNisto
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Homework Statement


i put my answers to the questions in bold

A spaceship and its occupants have a total mass of 1.8×105kg . The occupants would like to travel to a star that is 30 light-years away at a speed of 0.70c. To accelerate, the engine of the spaceship changes mass directly to energy.

How much mass will be converted to energy to accelerate the spaceship to this speed?

Assume the acceleration is rapid, so the speed for the entire trip can be taken to be 0.70c, and ignore decrease in total mass for the calculation. How long will the trip take according to the astronauts on board?

Homework Equations


The Attempt at a Solution


part a)
substituting the numebrs into the first equation
√(1-(0.70)2) = .7141428429
1/.7141428429 = 1.400280084
1.400280084 - 1 = .400280084moc2
.40(180000kg) = 72000kg

part b
t = d/v
(30y)c/0.70c = 42.9 y

42.9y = tΔ/√(1-(0.70)2) = tΔ0 = 30.6y
 
Last edited by a moderator:
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Please include units in your calculations, otherwise they are totally meaningless. (I'm not just being pedantic here: it's actually impossible to know what the calculations mean without units).

You have the right basic idea, but I think you need to be a little more careful. You have assumed that the mass of the ship is constant, when in fact it changes. The equation for the relativistic energy (rest energy PLUS kinetic) is indeed$$E = \gamma mc^2$$where ## \gamma = (\sqrt{1-v^2/c^2})^{-1}##. So, the difference in energy between moving (gamma is [STRIKE]non zero[/STRIKE] greater than 1 ) and not moving (gamma is [STRIKE]zero[/STRIKE] 1) would indeed be ##\Delta E = KE = (\gamma-1)mc^2## IF the ship's mass were not changing. BUT the ship's mass is changing, so the difference in energy is actually ##\Delta E = \gamma m_1c^2 - m_0c^2## where m1 is the mass after reaching 0.7c and m0 is the original mass. So, the algebra to solve for m1 here is a little more involved.
 
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i have included the units i hope that helps

cepheid said:
Please include units in your calculations, otherwise they are totally meaningless. (I'm not just being pedantic here: it's actually impossible to know what the calculations mean without units).

You have the right basic idea, but I think you need to be a little more careful. You have assumed that the mass of the ship is constant, when in fact it changes. The equation for the relativistic energy (rest energy PLUS kinetic) is indeed$$E = \gamma mc^2$$where ## \gamma = (\sqrt{1-v^2/c^2})^{-1}##. So, the difference in energy between moving (gamma is non zero) and not moving (gamma is zero) would indeed be ##\Delta E = KE = (\gamma-1)mc^2## IF the ship's mass were not changing. BUT the ship's mass is changing, so the difference in energy is actually ##\Delta E = \gamma m_1c^2 - m_0c^2## where m1 is the mass after reaching 0.7c and m0 is the original mass. So, the algebra to solve for m1 here is a little more involved.
 
AyooNisto said:
i have included the units i hope that helps

Okay...but did you read any of the rest of my post? I pointed out that you were doing something wrong. You need to correct it.

Also, I edited my first post to correct a stupid mistake about the gammas. I did so transparently.
 
AyooNisto said:
part a)
substituting the numebrs into the first equation
√(1-(0.70)2) = .7141428429
1/.7141428429 = 1.400280084
1.400280084 - 1 = .400280084moc2
.40(180000kg) = 72000kg
You have calculated the final kinetic energy of the spaceship, given that the mass of the spaceship has not changed. But as cepheid says, they want you to do the calculation given that the rest mass of the spaceship is used up to provide the speed increase. I think the best way to do the problem is to think of what they want you to assume stays the same in this problem. (Usually it is the rest mass that stays the same, but this is not true in this problem).
 
cepheid said:
the difference in energy is actually ##\Delta E = \gamma m_1c^2 - m_0c^2## where m1 is the mass after reaching 0.7c and m0 is the original mass. So, the algebra to solve for m1 here is a little more involved.
Isn't it quite a bit more involved? With conventional propellants, you have to consider how the mass changes over time, yielding a logarithmic relationship. Doesn't something similar apply here?
E.g., if you think about the momentum viewed from an inertial frame, and the acceleration comes from emitting radiation, as the craft accelerates the wavelength increases for the observer, reducing the momentum per photon.
 
since the question doesn't give an exhaust velocity or specific impulse, I think we are meant to assume that all the energy is contained in either the rest mass of the ship or the KE of the ship. In other words, there is no propellant and momentum is not conserved at all. In this case, it is not very involved, but maybe kindof useless, since in real problems we will always have conservation of momentum.

edit: maybe he is meant to assume that the 'propellant' is just EM radiation. Then it would be possible to solve the problem (with conservation of momentum). But this would take some time to do from first principles... more time than I would expect from the way the question is worded. So I don't think this is what he is meant to do.
 
BruceW said:
since the question doesn't give an exhaust velocity
I was trying to edit my post, but it got fouled up and I deleted it to I create another post.

The first part of the post asks about the amount of mass required to accelerate the ship, ingoring mass loss. So it seems it's asking for how much mass if converted into energy equals the kinetic energy of the space ship. It's not clear if energy added to the "exhaust" plume is to be considered.

The second part of the post is asking about time, how long will the trip take at 0.70 C, apparently from a reference frame of the space ships initial velocity, which would make it's intial speed zero, before it accelerated to 0.70 C.
 
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rcgldr said:
The first part of the post asks about the amount of mass required to accelerate the ship, ingoring mass loss. So it seems it's asking for how much mass if converted into energy equals the kinetic energy of the space ship. It's not clear if energy added to the "exhaust" plume is to be considered.
ah... that looks like what the OP did. But I interpreted the question differently. I essentially interpreted it as "what is the required change in rest mass of the spaceship, such that it can get to that velocity, given that total energy is constant (no energy is 'added'), and assuming no propellant, i.e. momentum not conserved"

I guess either interpretation is possible... so maybe I shouldn't have been so quick to disagree with the OP's original answer.
 
  • #10
mass doesn't increase, the correct phenomena is impedance aka relative mass. Just puting it out there since many people get confused with mass (the amount of matter in an object) and impedance (inertia; relative mass) which are both used interchengably as mass.
 
  • #11
Not necessarily. the rest-mass stays constant only if a pure four-force acts on the object. Otherwise, generally the rest-mass will change. And by the way part (b) in the question is worded, it looks like the rest-mass is supposed to change in part (a)
 

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