The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as anobject moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression forthe acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius R ofthe Earth, g, and h.&h=Suppose a 74.35 kg hiker has ascended to a height of 1893 m above sea level in the process of climbing Mt. Washington. Bywhat percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Useg = 9.807 m/s² and R = 6.371 × 106 m. Enter your answer as a positive value.weight change =%

Answered: The acceleration due to gravity, g, is…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter4: Motion In Two And Three Dimensions

Section: Chapter Questions

Problem 83AP: An elephant is located on Earth’s surface at a latitude . Calculate the centripetal acceleration of...

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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 Earth's 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 R of
the Earth, g, and h.
&h=
Suppose a 74.35 kg hiker has ascended to a height of 1893 m above sea level in the process of climbing Mt. Washington. By
what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use
g = 9.807 m/s² and R = 6.371 × 106 m. Enter your answer as a positive value.
weight change =
%

Expert Solution

Step 1

We know that acceleration due to gravity (denoted by "g") is the force of attraction between two objects, usually the Earth and an object near its surface. It is a constant value at any given point on the Earth's surface, approximately $9.8 m / s^{2} .$

So, g = $\frac{G M}{R^{2}}, G i s g r a v i t a t i o n a l c o n s \tan t, M i s m a s s o f e a r t h, a n d R i s r a d i u s o f e a r t h .$

However, as an object is lifted higher above the Earth's surface or sea level, the acceleration due to gravity decreases. This is because the gravitational force between the object and the Earth becomes weaker as the distance between them increases.

The relationship between the acceleration due to gravity and the height , h above the sea level can be described using the following equation:

$g_{h}$ = $\frac{G M}{{(R + h)}^{2}}$

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