Question

At what altitude above the earth's surface would the acceleration due to gravity be
4.9ms−2
? Assume the mean radius of the earth is
6.4×106
metres and the acceleration due to gravity
9.8ms−2
on the surface of the earth

Answer 1

Answer on Question#51974 - Physics - Other

At what altitude above the earth's surface would the acceleration due to gravity be $g_{h} = 4.9\frac{\mathrm{m}}{\mathrm{s}^{2}}$? Assume the mean radius of the earth is $R_{E} = 6.4 \times 10^{6}$ meters and the acceleration due to gravity $g_{0} = 9.8\frac{\mathrm{m}}{\mathrm{s}^{2}}$ on the surface of the earth.

Solution:

The acceleration due to gravity at some altitude $H$ above the earth's surface is given by

$$g_{H} = \frac{G M_{E}}{(R_{E} + H)^{2}},$$

where $M_{E}$ – is the mass of the earth, $G$ – is the gravitational constant.

It is given that

$$g_{h} = \frac{G M_{E}}{(R_{E} + h)^{2}} = 4.9 \frac{\mathrm{m}}{\mathrm{s}^{2}},$$

and that

$$g_{0} = \frac{G M_{E}}{R_{E}^{2}} = 9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}.$$

Dividing $g_{0}$ by $g_{h}$ we obtain

$$\frac{g_{0}}{g_{h}} = \frac{(R_{E} + h)^{2}}{R_{E}^{2}} = \left(1 + \frac{h}{R_{E}}\right)^{2}$$

Therefore

$$h = R_{E} \left(\sqrt{\frac{g_{0}}{g_{h}}} - 1\right) = 6.4 \times 10^{6} \mathrm{m} \left(\sqrt{\frac{9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}}{4.9 \frac{\mathrm{m}}{\mathrm{s}^{2}}}} - 1\right) = 2.65 \times 10^{6} \mathrm{m}$$

**Answer**: $h = R_{E} \left( \sqrt{\frac{g_{0}}{g_{h}}} - 1 \right) = 2.65 \times 10^{6} \mathrm{m}$.

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