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Imagine that the Earth's size shrunk by a factor of 5.70 from a radius of 1120 km to a radius of 6370 km. However, Earth's mass has remained the same at 5.98×1024 kg. If a person's mass is 80.7 kg, calculate their weight before and after Earth's shrinkage in Newtons and pounds

Respuesta :

The weight of the person in the first case is 25759.44 N while in the second it is 790.86 N.

What is weight?

Weight is the product of mass and acceleration due to gravity. We have to find the acceleration due to gravity in both cases.

g = GM/r^2

g1 = 6.67 × 10^−11 m3 kg−1 s−2 * 5.98×10^24 kg/(1120 * 10^3 m)^2

g1 = 3.99 * 10^14/1.25 * 10^12

g1 = 319.2 m/s^2

g2 = 6.67 × 10^−11 m3 kg−1 s−2 * 5.98×10^24 kg/( 6370  * 10^3 m)^2

g2 = 3.99 * 10^14/4.06 * 10^13

g2 = 9.8 m/s^2

The weight of the person in the first case is;  80.7 kg * 319.2 m/s^2

= 25759.44 N

The weight of the person in the second case is; 80.7 kg *  9.8 m/s^2

= 790.86 N

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