The acceleration due to gravity, g , is given by (c) What is the percentage chan

Question

The acceleration due to gravity,g, is given by

(c)What is the percentage change ingwhen moving from sea level to the top of Mount Elbert (a mountain over 14,000 feet tall in Colorado; in km, its height is 4.29 km; assume the radius of the Earth is 6400 km)?
percent change =_____?

The acceleration due to gravity, g , is given by g= r^2 GM The acceleration due to gravity ,g, is given by g Where M is the mass of the Earth, r is the distance from the center of the Earth, and G is the uniform gravitational constant. (a)Suppose that we increase from our distance from the center of the Earth by a distance The acceleration due to gravity, g, is given by gr=x. Use a linear approximation to find an approximation to the resulting change in g, as a fraction of the original acceleration: The acceleration due to gravity, g, is given by gg The acceleration due to gravity, g, is given by gg The acceleration due to gravity ,g, is given by g______? (c)What is the percentage change in g when moving from sea level to the top of Mount Elbert (a mountain over 14,000 feet tall in Colorado; in km, its height is 4.29 km; assume the radius of the Earth is 6400 km)? percent change =_____?

Explanation / Answer

g = (GM)/r^2 dg/dr = -2 (GM) / r^3 dg = -2(GM)/r^3 dr That is it. If that is not what you want, please post the actual problem and we can advise from the start. You might not be needing this differential. ------ You do realize that it does not matter if constants are squared (GM), you still have r^{-2} which differentiates to -2 r^{-3}, which is what all of us wrote (and we never squared the whole thing either...I do not understand what you are talking about). (a) g(r) = GM/r^2 taking the distance to be Delta r = x, we have g(r + x) = GM / (r + x)^2 = GM / { r^2 (1 + x/r)^2} The quantity x/r

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