Question

A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is $0.0075m/s^2$ less than that at sea level. What is the observatory’s altitude?

Short answer

The observatory’s altitude is$2.44km.$

Step-by-step solution

Given information.

The free-fall acceleration of observatory is $0.0075m/s^2$less than that at sea level.

Calculation.

Consider the height of the observatory is h.

Hence, the free-fall acceleration on the observatory is$g=\mathrm{GM}(\mathrm{R}+\mathrm{h})2.$

So, the difference in the free-fall acceleration is :

$$\begin{gathered} \Delta \mathrm{g}=\mathrm{GMR}2-\mathrm{GM}(\mathrm{R}+\mathrm{h})2 \\ =\mathrm{GMR}2-\mathrm{GMR}2[1+\mathrm{hR}]2. \\ =\mathrm{GMR}2[1-(1+\mathrm{hR})-2]. \\ =\mathrm{GMR}2[1-1+2\mathrm{hR}]. \end{gathered}$$

Ignoring higher order term as $h\ll R=2ghR.$

Now, rearrange the above equation to get the h,

$\mathrm{h}=\Delta \mathrm{gR}2\mathrm{g}.$

Now substitute the values of g, R and $\Delta g$to calculate the

localid="1648481629645" $$\begin{gathered} \mathrm{h}=(0.0075\mathrm{m}/{\mathrm{s}}^2)(6.37\times {10}^6\mathrm{m})2(9.8\mathrm{m}/{\mathrm{s}}^2) \\ =2.44\times {10}^3\mathrm{m} \\ =2.44\mathrm{km}. \end{gathered}$$

Final answer.

The observatory’s altitude is$2.44km.$