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 x 106 metres and the acceleration due to gravity 9:8ms-2 on the surface of the earth

Answer 1

At what altitude above the earth's surface would the acceleration due to gravity be $4.9 \, \text{ms}^{-2}$? Assume the mean radius of the earth is $6.4 \times 10^{6}$ metres and the acceleration due to gravity 9.8 $\text{ms}^{-2}$ on the surface of the earth.

Solution.

On the Earth surface, the gravitational force, acting on the point-like constant mass object, is described by the following equation:

$$F (R _ {E}) = G \frac {m M _ {E}}{R _ {E} ^ {2}},$$

where $G$ is the gravitational constant, $m$ is the mass of a man and $M$ is the mass of the Earth. Dividing it to the mass $m$, we obtain

$$\frac {F (R _ {E})}{m} = g (R _ {E}) = G \frac {M _ {E}}{R _ {E} ^ {2}}.$$

Gravitational accelerations at different altitudes are than connected in the following way:

$$g (R _ {E} + h) = g (R _ {E}) \frac {R _ {E} ^ {2}}{(R _ {E} + h) ^ {2}}.$$

Solving equation, we find the altitude $h$:

$$h = R \left(\sqrt {\frac {g (R _ {E})}{g (R _ {E} + h)}} - 1\right),$$ $$h = 6.4 \times 10^{6} \mathrm{m} \left(\sqrt {\frac {9.8 \, \mathrm{ms}^{-2}}{4.9 \, \mathrm{ms}^{-2}}} - 1\right) = 6.4 \times 10^{6} \mathrm{m} \times 0.414 = 2.7 \times 10^{6} \mathrm{m}.$$

Answer.

$$h = 2.7 \times 10^{6} \mathrm{m}$$

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