Unit 3 - Problem 3 => Gravity on Mt. Everest

Given:

• $g_{sea-level} = \frac{10\;meters}{seconds^2}$
• Gravity is inversely proportional to the square of the distance from the center of the earth.
• The earth's radius is aproximately 6,400 km.
• Mt. Everest height is about 9 km.
• Calculate constant of proportionality k.
• Calculate by what percentage does the acceleration due to gravity change when we stand on Mt. Everest.

Rational for constant of proportionality:

• $g = \frac{k}{r^2}$

Calculating the constant of proportionality:

• $10 = \frac{k}{6400^2}$
• $k = 10*6400^2$

Rational for finding the acceleration due to gravity when we stand on Mt. Everest.

• Mt. Everest is about 9 km high so its distance from the center of the earth is about 6409 km.
• $g_{Mt.Everest} = \frac{k}{r^2} = \frac{10*6400^2}{r^2}$

Calculating Mt.Everest force of gravity

• $g_{Mt.Everest} = \frac{10*6400^2}{r^2}$
• $g_{Mt.Everest} = \frac{10*6400^2}{6409^2} = 9.971934215130506 = 9.97$

Rational for percent difference:

• ${percent-difference} = 100 \times \left(\frac{Difference}{{Comparison Value}}\right)$

Calculating the % difference in force of gravity at sea level vs Mt.Everest

• ${percent-difference} = 100 \times \left(\frac{g_{sea-level}-g_{Mr.Everest}}{g_{sea-level}}\right)$
• ${percent-difference} = 100*\left(\frac{(10-9.97)}{10}\right)= 0.2999999999999936 = 0.3\;\%$