At what height above the surface of Earth is there a 1% difference between the approximate magnitude of gravitational field 9.8Nkgand the actual value of gravitational field at that location? This is, at what height y above the Earth’s surface isGMe(Re+y)2=0.99GMeRe2

Short Answer

Expert verified

At a height of 32 Km above the surface of Earth, there is a 1% difference between the approximate magnitude of the gravitational field and the actual magnitude of the gravitational field at that location.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • Assume, the radius of the earth is,Re=6400km
02

Significance of gravitational law of motion

The gravitational force formula defines the magnitude of the force between the two objects. It is also known as Newton’s law of gravitation.

Gravitational force = (Gravitationalconstant)(massofobject1)(massofobject2)(distancebetweenthetwoobjects)

03

Determination of height above the surface of the earth

The value of G at a height “h” above the earth’s surface from the gravitational law of motion can be expressed as,

gh=GMe(Re+h)2

Here, G is the gravitational constant,Me is the mass of the earth,Re is the radius of Earth.

Rewriting the above equation,

gh=GMeRe1+hRe2gh=gs1+hRe2


forh<<Re

role="math" localid="1658907029410" gh=gs1+hRe-2=gs1-2hRegh=gs1-2hRegh=gs1-1100gh=0.99gs

Therefore,

0.99=1-2hRe

Now, Re=6400km

Substitute all the values in the above equation.

2hRe=0.01h=0.01×64002h=32km

Thus, at height 32 km above the surface of Earth, there is a 1% difference between the approximate magnitude of gravitational field and actual magnitude of gravitational field at that location

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