ASTRONOMY FUNDAMENTALS - Chapter 4, Exercise 2 Solution
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Scaling the Sun's Fusion Rate to One Day

PROBLEM
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The Sun's core fuses approximately 600 billion kilograms of hydrogen
into helium every second. Calculate approximately how much hydrogen the
Sun fuses in one full day (24 hours), expressed in scientific notation.

SOLUTION
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Step 1 - find the number of seconds in one day:
    24 hours x 60 minutes/hour x 60 seconds/minute = 86,400 seconds

Step 2 - multiply the per-second fusion rate by the number of seconds
in a day:
    600,000,000,000 kg/second x 86,400 seconds
    = 600,000,000,000 x 86,400 kg
    = 51,840,000,000,000,000 kg

Step 3 - express in scientific notation:
    51,840,000,000,000,000 kg = 5.184 x 10^16 kg

ANSWER: The Sun fuses approximately 5.184 x 10^16 kilograms of hydrogen
in one day - roughly 5.18 x 10^13 tonnes (metric tons).

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WHY THIS WORKS AS AN ANSWER

Because the chapter's own figure (600 billion kg/second) is a rate -
an amount per unit of time - finding the total amount fused over a
longer period is simply a matter of multiplying that rate by the
number of matching time units in the period being asked about. Since
the rate is given per second, the day first needs to be converted into
seconds (24 x 60 x 60 = 86,400) before the multiplication makes sense.

This result also helps put the chapter's own "more than ten billion
years to fully convert the Sun's core hydrogen" claim into a more
concrete perspective: even at this genuinely enormous daily fusion
total - over 50 quadrillion kilograms every single day - the sheer
total mass of hydrogen in the Sun's core is large enough that it would
still take billions of years to use it all up at this rate.