Acceleration Due to Gravity Video Lessons

Concept

# Problem: Suppose the free-fall acceleration at some location on earth was exactly 9.8000 m/s2. What would it be at the top of a 1000m -tall tower at this location? (Ans. in 5 sig.figs)

###### FREE Expert Solution

Acceleration due to gravity, g:

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

At the height, h, the acceleration due to gravity will be:

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

Dividing gI by g:

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###### Problem Details

Suppose the free-fall acceleration at some location on earth was exactly 9.8000 m/s2. What would it be at the top of a 1000m -tall tower at this location? (Ans. in 5 sig.figs)