Solve the following :


What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of $8848 \mathrm{~m}$. The value at sea level is $9^{\prime} 80 \mathrm{~m} / \mathrm{s}^{2}$.


$g^{\prime}=g\left(1-\frac{2 \hbar}{R}\right)$

g' be the acceleration due to gravity at Mount Everest


$g^{\prime}=9.8 \times\left(1-\frac{2 \times 8848}{6400 \times 10^{\mathrm{s}}}\right)$

$=9.77 \mathrm{~m} / \mathrm{s}^{2}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now