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 #40852, Physics, Mechanics

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

Solution:

The velocity of a freely falling body increased at a steady rate i.e., the body had acceleration. This acceleration is called acceleration due to gravity $g$ .

Let a body of mass $m$ be placed on the surface of the Earth:

$$g = G \frac {M}{R ^ {2}}$$

where $M$ is the mass of the Earth, $R$ is the radius of the Earth and $G$ is the gravitational constant.

et the body be now placed at a height $h$ above the Earth's surface. Let the acceleration due to gravity at that position be $g'$ .

Then,

$$g ^ {\prime} = G \frac {M}{(R + h) ^ {2}}$$

For comparison, the ratio between $g'$ and $g$ is taken

$$\frac {g ^ {\prime}}{g} = G \frac {M}{(R + h) ^ {2}} \frac {R ^ {2}}{G M} = \left(\frac {R}{R + h}\right) ^ {2}$$

Thus,

$$h = R \left(\sqrt {\frac {g}{g ^ {\prime}}} - 1\right)$$ $$h = 6. 4 \cdot 1 0 ^ {6} \cdot \left(\sqrt {\frac {9 . 8}{4 . 9}} - 1\right) = 2. 6 5 \cdot 1 0 ^ {6} \mathrm {m}$$

Answer. $h = 2.65 \cdot 10^{6} \, \text{m}$ .