A spaceship of mass 2.55 106 kg is to be accelerated to a speed of 0.645c. (a) W

Question

A spaceship of mass 2.55 106 kg is to be accelerated to a speed of 0.645c. (a) What minimum amount of energy does this acceleration require from the spaceship's fuel, assuming perfect efficiency? (b) How much fuel would it take to provide this much energy if all the rest energy of the fuel could be transformed to kinetic energy of the spaceship? i've solved part A but it says it's completely wrong. and part b, i've no clue how to solve it. part A this is how i did. (0.645c)(2.55e6)(3e8)^2= 1.48e23

Explanation / Answer

Actually, the speed is closer to the speed of light.. so the mass of the space craft will not be m ...

mass of spacecraft = m / sqrt ( 1 - v^2 / c^2 )

where.. v = velocity of space craft .. here 0.645 c

so... mass = 2.55*10^6 / sqrt ( 1 - (0.625)^2 ) = (2.55/0.78062475) * 10^6 = 3.266614 * 10^6 Kg

so.. final energy = 0.5* 3.266614*10^6 * (0.625 * 3 * 10^8 )^2 = 5.167886 * 10^23 Joules

b)

Let teh mass of fuel be m..

rest mass energy = m * c^2 = 5.167886 * 10^23

so.. m * ( 3 * 10^8)^2 = 5.167886 * 10^23

so rest mass of fuel = 5.7421 * 10^6 kg of fuel

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