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At what distance above the surface of the earth is the acceleration due to the earth鈥檚 gravity 0.980 m/s2 if the acceleration due to gravity at the surface has magnitude 9.8 m/s2?

Short Answer

Expert verified

Answer

At a distance of 1.38 x 107 m above the Earth's surface, the acceleration due to gravity will be 0.980 m/s2 .

Step by step solution

01

Step-by-Step Solution Step 1: Identification of given data

  • The acceleration due to gravity at the unknown distance is g鈥=0.980 .
  • The acceleration due to gravity at the surface of Earth is g=9.80 .
02

Step-2: Estimation of the distance from the surface of Earth for given acceleration due to gravity

The relation between g and g鈥 is,

g'=g10

The acceleration due gravity above a distance h from the surface of Earth can be expressed as,

g'=GMR+h2

Here, g' is the acceleration due to gravity at height h from the surface of Earth M is the mass of the Earth, G is the gravitational constant, R is the radius of Earth.

Substitute the value for g' in the above equation,

g10=GMR+h2(i)

The acceleration due to gravity g can be expressed as,

g=GMR2(ii)

Compare the equations (i) and (ii),

GM10R2=GMR+h210R2=R+h2

Evaluate the value of h from the above equation,

10R2=R+h210R=R+hh=R101

Radius of Earth isR=6.37106鈥尘. Thus, the distance h is calculated as,

h=6.37106鈥尘101h=1.38107鈥尘

Thus, the acceleration due to gravity will be 0.980 m/s2 at a distance of 1.38 x 107m above the Earth's surface.

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