Question

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 m. The value at sea level is 9.80 m s−2.

Answer 1

Explanation:

Given,

h=8848mh=8848m

The value of sea level is 9.08m/s^29.08m/s

2

So,

Let g'g

′

be the acceleration due to the gravity on the Mount Everest.

g'=g(1-\dfrac{2h}{h})g

′

=g(1−

h

2h

)

=9.8(1-\dfrac{17696}{6400000})=9.8(1−

6400000

17696

)

=9.8(1-0.00276)=9.8(1−0.00276)

9.8\times0.997249.8×0.99724

=9.773m/s^2=9.773m/s

2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2=9.773m/s

2

Answer 2

Explanation:

• The acceleration due to gravity on the top of Mount Everest can be defined as $g^{\prime}=g\left(1-\left(\frac{2 h}{R}\right)\right)$ where g is the acceleration due to gravity at sea level.  
• It is given that the height of the world’s highest mountain peak, h = 8848m. Also, the acceleration due to gravity at sea level is given as 9.80 m/s².  
• Thus, on substituting the known values, we get, $g^{\prime}=9.8(1-0.00276)=9.77 \mathrm{m} / \mathrm{s}^{2}$.
• Therefore, the acceleration due to gravity on the top of Mount Everest, $g^{\prime}=9.77 \mathrm{m} / \mathrm{s}^{2}$ .