Question

At what height above the earth’s surface would the acceleration due to gravity be 4.9m/s2? Assume the mean radius of the earth is 6.4x106m and the acceleration due to gravity on the earth surface is 9.8m/s2

Answer 1

Answer on Question 55212, Physics, Mechanics | Kinematics | Dynamics

Question:

At what height above the Earth's surface would the acceleration due to gravity be $4.9\,m/s^2$? Assume the mean radius of the Earth is $6.4 \cdot 10^6\,m$ and the acceleration due to gravity on the Earth surface is $9.8\,m/s^2$.

Solution:

Let a body of mass $m$ be placed on the Earth surface, whose mass is $M$ and radius is $R$. Then, the acceleration due to gravity on the Earth surface looks like:

$$g = \frac{GM}{R^2}.$$

Now we placed the body at a height $h$ above the Earth's surface. Then, the acceleration due to gravity above the Earth's surface $g'$ looks like:

$$g' = \frac{GM}{(R + h)^2}.$$

Let's take the ratio between $g$ and $g'$:

$$\frac{g}{g'} = \frac{(R + h)^2}{R^2} = \frac{9.8\,m/s^2}{4.9\,m/s^2} = 2.$$

Then, we get quadratic equation from which we can obtain $h$:

$$(R + h)^2 = 2R^2,$$ $$R^2 + 2Rh + h^2 = 2R^2,$$ $$h^2 + 2Rh - R^2 = 0.$$

This equation has two roots:

$$D = b^2 - 4ac = (2R)^2 - 4 \cdot 1 \cdot (-R^2) = 4R^2 + 4R^2 = 8R^2,$$ $$h_1 = \frac{-b - \sqrt{D}}{2a} = \frac{-2R - \sqrt{8R^2}}{2} = \frac{-2R - 2R\sqrt{2}}{2} = -R - R\sqrt{2}.$$ $$h_2 = \frac{-b + \sqrt{D}}{2a} = \frac{-2R + \sqrt{8R^2}}{2} = \frac{-2R + 2R\sqrt{2}}{2} = -R + R\sqrt{2}.$$

Because the height can't be negative the correct answer is $h = -R + R\sqrt{2}$.

Then, we can calculate the height above the Earth's surface at which the acceleration due to gravity would be $4.9\,m/s^2$:

$$h = R \sqrt {2} - R = R \left(\sqrt {2} - 1\right) = 0.41R = 0.41 \cdot 6.4 \cdot 10^{6}m = 2.624 \cdot 10^{6}m.$$

Answer:

$$h = 2.624 \cdot 10^{6}m.$$

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