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A 85 kg man weighs 833 N on the Earth's surface. How far above the surface of the Earth would he...

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An 85 kg man weighs 833 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 18% of his body weight?

Gravitation pull of Earth

The weight given as {eq}w = mg {/eq} where {eq}m {/eq} is mass and {eq}g {/eq} is the acceleration due to gravity is just the gravitational pull of Earth at the surface on the mass {eq}m {/eq}. In general, the weight of a body of mass {eq}m {/eq} a distance {eq}R {/eq} from the center of Earth is

{eq}w = \dfrac{GM_E m}{R^2}, {/eq}

where {eq}G {/eq} is the gravitational constant and {eq}M_E {/eq} is the mass of Earth. We retrieve {eq}w = mg {/eq} when {eq}R = R_E {/eq} (Earth's radius), i.e. {eq}g = GM/R_E^2 {/eq}.

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To lose a body weight of 18% means that if {eq}w {/eq} is the weight at the surface {eq}R_E {/eq}, then {eq}w' = (1-.18)w {/eq} at some distance...

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Gravitational Pull of the Earth: Definition & Overview

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Chapter 15 / Lesson 17
29K

Earth's gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we'll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.


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