Answer to Question #105229 in Classical Mechanics for Jeff

Question

In most calculations, we use W \approx mgW≈mg. But we know that, for large changes in altitude, we need to use W∝1/r2. How far above the Earth's surface can we use W = mgW=mg before our systematic error reaches 1%? Use only the information given in this question, and the radius of the Earth re=6400 km. Do not explicitly use GG or the mass of the Earth, and do the calculation for the pole so that we don't worry about the effect of centripetal acceleration.

Hint: Does gg become larger or smaller with altitude?

Altitude= ___ \text{km}km (to one significant figure)

Answer 1

The acceleration due to gravity becomes smaller with altitude, and here is why. At the earth's surface:

$$G\frac{Mm}{r_e^2}=mg_0,\\ g_0=\frac{GM}{r_e^2}.$$

At some high altitude above the surface of the earth:

$$G\frac{Mm}{(r_e+h)^2}=mg_h,\\ g_h=\frac{GM}{(r_e+h)^2}.$$

See: when the height increases, g decreases.

Our error must be at least 0.01, or mathematically

$$\epsilon=1-\frac{g_h}{g_0},\\ \space\\ g_h=(1-\epsilon)g_0=(1-\epsilon)\frac{GM}{r_e^2}.$$

On the other hand:

$$\frac{GM}{(r_e+h)^2}=(1-\epsilon)\frac{GM}{r_e^2},\\ \space\\ (r_e+h)^2=\frac{r_e^2}{(1-\epsilon)},\\ \space\\ h=r_e\bigg(\frac{1}{\sqrt{1-\epsilon}}-1\bigg).$$

The last equation gives 32 (30 to one significant figure) km for 1% difference with the accepted value.

Interestingly, the acceleration due to gravity like on the Moon (1.62 m/s²) will be at the altitude of 9350 km above the earth's surface.