How do you apply E=MC^2?

Question

I was wondering how you would apply E=MC^2 to a physics problem. The problem is asking for the maximum mass of a rocket that can reach an altitude of 345 km. It can convert 1 g matter into thrust energy by E+MC^2. I admit this is a homework problem so you don't have to give me the answer but just simply how to start working it. My teacher did not go over this and I can't find it in my book. Thank you in advance.

Sorry I meant E=MC^2 in the question.

Answer 1

The amount of energy created when you converter mass into energy is equal the product of the lost mass time the speed of light squared.

So this means if you could take a kg of anything and turn it into energy then you would release 90,000,000,000,000,000 Jules of energy.

To find the answer to your problem find the energy released by converting 1g into pure energy. Then find the potential energy of a rocket of mass m at 345km above the earth. Set the two energy's equal to each other and solve for m.

Answer 2

You solve this problem using energy conservation, and recognizing the different forms of energy involved. The first is the potential energy of the rocket: this is given by E=mgh, where m= rocket's mass, h= altitude and g = accel of gravity. This is also how much kinetic energy needed to lift the rocket to altitude h. All the energy is being supplied by the conversion of 1g of matter into energy. Call the rocket's mass Mr, the "fuel" mass Mf (= 1 gm) we get:

h = E/Mr*g = Mf*c^2/Mr*g

Keep the units straight (g=m/sec^2, h = m c = 3*10^8 m/sec)

EDIT: The equation above is for the altitude, but you want the mass of the rocket Mr

Mr = E/h*g = Mf^c^2/h*g

This comes out 2.7*10^7kg

Note: at the altitude you were given, the variation in g is very small:
h^2/Re^2, where Re is the earth radius. This is 0.3%, small enough to ignore and avoid the integration.

Answer 3

Assume that the maximum mass of the rocket is way greater than the 1 g mass it will use to propel it.

The max mass of the rocket will be computed by equating the work done in moving the rocket from the surface of the earth to the 345 km altitude.

Bear in mind that the weight of the rocket varies with its distance from the earth, this means that the acceleration due to gravity is not constant. The acceleration due to gravity at any height is given by:

a = g (r/s)^2

where g=9.8 m/s^2
r = radius of the earth
s=height from the center of the earth or radius plus altitude.

This relation is intuitive because the force of gravity on a body is inversely proportional to its distance from the center of another heavenly body. And force is directly proportional to acceleration.

The work done is then given by:

integral of [ m g (r/s)^2 ] ds evaluated from r to 345 km

(you must know your calculus then)

where m = mass of the rochet
the term in brackets is the force on the rocket at any altitude
ds is the infinitesimal distance of the flight

Equate this integral with the energy that the 1 g mass will release.

As you can see the mass of the rocket can be factored out because we have assumed it to be way greater than the fuel. A rigorous approach would be to include the mass of the fuel, relating it with the altitude. Please adopt a consistent set of units in your work.

Answer 4

Actually...
This equation was used in the development of the atomic bomb. By measuring the mass of different atomic nuclei and subtracting from that number the total mass of the protons and neutrons as they would weigh separately, one could obtain an estimate of the binding energy available within an atomic nucleus. This could be and was used in estimating the energy released in the nuclear reaction, by comparing the binding energy of the nuclei that enter and exit the reaction.

It is a little known piece of trivia that Einstein originally wrote the equation in the form Δm = L/c² (with an "L", instead of an "E", representing energy, the E being utilised elsewhere in the demonstration to represent energy too).

Answer 5

The thrust energy is only part of the problem. The real thing that you need to know is the specific impulse of the engine. This is defined as the time, in seconds, that one pound of fuel can create one pound of thrust. (For present-day chemical rocket engines, 250 to 300 seconds is par for the course.) Where we are going with this is to attempt to determine the momentum (per unit time) of the exhaust, which is the product of its mass and velocity. The energy per unit time (power) is the product of mass and velocity squared. So, e = m c squared gives you an energy, but without knowing some additonal fact (such as the exhaust velocity) you don't have anything. You can use Einstein to calculate the energy available from a particular form of fuel; if for example, you are converting hydrogen (1.008) into helium (4.003), you get .029 amu of mass converted to energy per helium atom produced.

Answer 6

Energy is equal to Mass times the speed of light squared.

Answer 7

the equasion means that matter can be converted to energy and vise versa, so 1 g of matter would equal the speed of light squared