Help! The acceleration due to gravity, g. is constant at sea level on the Earth\

Question

Help!

The acceleration due to gravity, g. is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh, Express the equation in terms of the radius of the Earth RE, g and h. A 91.75 kg hiker has ascended to a height of 1.880 times 103 m in the process of climbing Ml Washington. By what percent has the hiker's weight changed by climbing to this elevation? Use g = 9.807 m/s2 and RE = 6.371 times 106m (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.) %W change =

Explanation / Answer

As I have already explained that acceleration due to gravity g is a variable quantity and it varies from place to place. We here, will derive the relation that describes variation of g with altitude.

Let us consider a body of mass M lying on the surface of earth of mass M and radius R. Let g be value of acceleration due to gravity on the free surface of earth.

Then

g = GM / R2 %u2026%u2026%u2026%u2026%u2026.. (i)

Suppose the body is taken to height %u2018h' above the surface of earth where the value ofacceleration due to gravity is gh.

Then

gh = GM / (R + h )2 %u2026%u2026%u2026%u2026(ii)

Where (R+h) is the distance between the centers of body and earth.

Dividing (ii) by (i), we get

gh / g = GM / (R + h )2 X R2 / GM = R2 / (R+ h ) 2

when h < < R, then

gh / g = R2 / R2 ( 1 + h2 / R2 ) 2

= 1/ ( 1 + h2 / R2 ) 2

= ( 1 + h2 / R2 ) -2

Since, h < < R, then h/r is very small as compared to I. Expanding the right hand side of the above equation by Binomial Theory and neglecting squares and higher powers of h/r, we get

gh / g = 1-2h / R

Thus, acceleration due to gravity decreases with increase in height / altitude.

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