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Radius of the earth is 6.4 × 10³ km and value of acceleration due to gravity on

its surface is 9.8 m/s². Find the value of acceleration due to gravity produced
on a meteor at a distance of 9850 m from the earth surface. (Ans: 9.76 m/s²)


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Answer:

solution

Step-by-step explanation:

The acceleration due to gravity produced by a planet or a celestial body at any point outside its surface is given by the formula:

a = G * M / r^2

where G is the universal gravitational constant (6.67 × 10^-11 N m^2 / kg^2), M is the mass of the celestial body, r is the distance between the center of the body and the point where the acceleration due to gravity is to be calculated.

In this problem, the distance between the meteor and the center of the Earth is:

r = radius of the Earth + 9850 m

r = 6.4 × 10^6 m + 9850 m

r = 6.41 × 10^6 m

The mass of the Earth is not given explicitly, but we can assume it to be 5.97 × 10^24 kg.

Substituting these values in the formula, we get:

a = G * M / r^2

a = (6.67 × 10^-11 N m^2 / kg^2) * (5.97 × 10^24 kg) / (6.41 × 10^6 m)^2

a = 9.76 m/s^2 (approx)

Therefore, the acceleration due to gravity produced on a meteor at a distance of 9850 m from the Earth surface is approximately 9.76 m/s^2.

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