Chapter 13: Q. 10 (page 353)

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

Short Answer

Expert verified

The observatory’s altitude is $2.44 k m .$

Step by step solution

Given information.

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

Calculation.

Consider the height of the observatory is h.

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

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

$Δ g = GMR 2 - GM (R + h) 2 = GMR 2 - GMR 2 [1 + hR] 2 . = GMR 2 [1 - (1 + hR) - 2] . = GMR 2 [1 - 1 + 2 hR] .$

Ignoring higher order term as $h ≪ R = 2 g h R .$

Now, rearrange the above equation to get the h,

$h = Δ gR 2 g .$

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

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

Final answer.

The observatory’s altitude is $2.44 k m .$

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

Short Answer

Step by step solution

Given information.

Calculation.

Final answer.

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A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0075m/s2 less than that at sea level. What is the observatory’s altitude?

Short Answer

Step by step solution

Given information.

Calculation.

Final answer.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

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Study anywhere. Anytime. Across all devices.

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