Question
Asked Sep 14, 2019
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Assuming that the Earth has a uniform density p= 5540.0 kg/m3, what is the value of the gravitational acceleration at a distance d= 4600 km from the earth's center? 

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Expert Answer

Step 1

Given information:

Density of earth (p) = 5540.0 kg/m3

Acceleration due to gravity at radius of (R) = 4600 km = 4600 × 103 m

Acceleration due to gravity (g) =?

Step 2

Let us consider a unit mass at the given radius of earth i.e. m = 1kg.

Let the mass of earth up to the given radius be (M), we can find M by:

M = D × V             (Where “V” volume of the earth till the given radius)

But,

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3 V 3 Substituting the corresponding values in the equation of M, we get the value of M as: 1020 kg M 4.07

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Step 3

According to newton law of gravitation the force be...

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Мт F G R2 тg Therefore, from the above equation we get "g" as: (6.67 x10 11m3kg1s-1)x4.07 x 1020 GM 0.0012 m/s2 R2 (4600 x 103)2

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