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2 years ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

Physics

1 answer:

shutvik [7]2 years ago

5 0

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

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