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Question: If the gravitational acceleration at the surface of earth is\[9.81m{s^{ - 2}}\]. What is its value a...

If the gravitational acceleration at the surface of earth is9.81ms−29.81m{s^{ - 2}}. What is its value at a height equal to the diameter of earth from its surface?

(A)4.905ms−1 (B)2.452ms−1 (C)3.27ms−1 (D)1.09ms−1  \left( A \right)4.905m{s^{ - 1}} \\\ \left( B \right)2.452m{s^{ - 1}} \\\ \left( C \right)3.27m{s^{ - 1}} \\\ \left( D \right)1.09m{s^{ - 1}} \\\
Explanation

Solution

We are first going to consider the value of the acceleration due to gravity as given, then, using the formula for the acceleration due to gravity at the surface of earth, one equation is obtained after that at the height equal to diameter of the earth, the acceleration due to gravity is found.

Formula used: The formula for the gravitational acceleration is given by
g=GMr2g = \dfrac{{GM}}{{{r^2}}}
Where, GGis the gravitational constant, MMis the mass of the earth and RRis the distance of that point

Complete step-by-step solution:
It is given that the gravitational acceleration at surface of earth is equal to
g=9.81ms−2g = 9.81m{s^{ - 2}}
The formula for the gravitational acceleration is given by
g=GMr2g = \dfrac{{GM}}{{{r^2}}}
Where, GGis the gravitational constant, MMis the mass of the earth and RRis the distance of that point
Putting the value of the gravitational acceleration, we get
9.81=GMR29.81 = \dfrac{{GM}}{{{R^2}}}
Now, at the point with height equal to diameter of earth that is twice the radius,
This means that now,
r=2Rr = 2R
Putting this in the equation for the acceleration due to gravity, we get
g′=GM(2R)2=14GMR2g' = \dfrac{{GM}}{{{{\left( {2R} \right)}^2}}} = \dfrac{1}{4}\dfrac{{GM}}{{{R^2}}}
Putting the value of GMR2\dfrac{{GM}}{{{R^2}}}in the above equation, we get
g′=14×9.81=1.09ms−2g' = \dfrac{1}{4} \times 9.81 = 1.09m{s^{ - 2}}
Thus, the value of acceleration due to gravity at a height equal to the diameter of earth from its surface is equal to1.09ms−21.09m{s^{ - 2}}.

Note: It is important to note that the acceleration of an object changes with altitude. The change in gravitational acceleration with distance from the center of Earth follows an inverse-square law. This relation has been used in order to find the acceleration at the different altitudes with respect to the earth.