Question

the energy output of the sun is approximately 4.0 x 10^26 J/s. If all this energy results from mass-energy conversion in the fusion process, calculate the rate at which the sun is losing mass.

Answer 1

I think we can approach this using differentials, which will introduce the rates for us.
We know that\[\huge E=mc^2\]Differentiating will yield:\[\huge \frac{dE}{dt}=c^2\frac{dm}{dt}\]You're given the rate of energy (should be negative because it's an output) and c is just the speed of light. Solve for the rate of mass (which should also be negative because it's losing mass)!

Answer 2

This is with respect to time, of course

Answer 3

Is there any way I could do this only using the E=mc^2 formula??

Answer 4

That is using the formula, just indirectly. The issue with using E=mc^2 is that those are exact instantaneous values. We're given RATES, which are not instantaneous. In order to calculate this, we have to find differentials which are also rates. To avoid the clutter, you can use:
\[\huge \text{E}'=c^2m'\]The ' (called "E prime" or "m prime") refers to the derivative/rates of each variable.

Answer 5

It's different notation, but the same meaning as what I wrote before. It just looks nicer.

Answer 6

M = E/c^2
E = 4.0 x 10^26 J = 4.0 x 10^26 / 10^-7 = 4.0 x 10^33 erg
c = 3 x 10^8 m/s^2

M = 4.0 x 10^33 / 9 x 10^16 = 4.4 x 10^16 g/sec

sun loses 4.4x10^16 grams per second

Answer 7

There is an issue with doing that. Mathematically it works out, given the linearity. Conceptually, E or m is not a rate -- they're given instantaneous values.

Answer 8

We are asked to find rates, so we must alter the equation so that our variables are also rates. We do that by differentiating. dm/dt can be spoken as "the rate of change of mass with respect to time" and similarly with dE/dt. This is what we want because this is what the problem asks for.

Answer 9

I think @IrishBoy123 can explain in better detail for you

Answer 10

thank you! I'm just a little confused because I haven't learned all this in class yet, the furthest I've learnt is E=mc^2

Answer 11

i'm with you on this @CShrix

hello @karlaLTR !!

Answer 12

Have you learned any sort of calculus yet?

Answer 13

not yet, I've only taking grade 11 physics at the moment

Answer 14

Mmm, this really what this problem requires. The curriculum might let you use E=mc^2 because the relationship between energy/mass and the rate of change of energy/mass is still linear.
E' = c^2 m'
E = c^2 m

That is why DariusX's method works.

Conceptually, this is wrong because mass and energy are not rates. They're specific given values. A rate is given by a change during a time interval. I think for your class' sake, you can just use E=mc^2. If you learn calculus later in HS/College, you'll see why this is like this.

Answer 15

I understand the logic in what you're saying, I think your way is a lot more efficient in terms or accuracy but It's easier to do it the other way for now since I'm not required to get too complicated for now, thank you for your help! You're really smart :)

Answer 16

You are very welcome X) Best of luck!