How far from the earth’s surface would a 72.1kg person experience a force of1.6 Newtons?
F = GMm/r^2
Look up G and M, and then plug in the numbers
The radius of the earth is 6356 km
Since F = GMm/r^2
Fr^2 = GMm and is constant, you want r such that
(1.6*r^2) = (72.1*9.8*6356^2)
Find r, then you want r-6356
To determine the distance from the Earth's surface where a 72.1kg person would experience a force of 1.6 Newtons, we can use Newton's law of universal gravitation:
F = (G * m1 * m2) / r^2,
where F is the force of gravity, G is the gravitational constant (approximately 6.67 x 10^-11 N(m/kg)^2), m1 and m2 are the masses of the two objects (in this case, the person's mass and the Earth's mass), and r is the distance between the centers of the two objects.
In this case, we know that the force experienced by the person is 1.6 Newtons and the person's mass is 72.1kg. The mass of the Earth is approximately 5.97 x 10^24 kg.
Rearranging the equation, we can solve for r:
r = √((G * m1 * m2) / F).
Plugging in the values, we get:
r = √((6.67 x 10^-11 N(m/kg)^2 * 72.1kg * 5.97 x 10^24 kg) / 1.6 N).
Calculating this, we find that the distance from the Earth's surface where a 72.1kg person would experience a force of 1.6 Newtons is approximately 1.23 x 10^7 meters, or 12.3 million meters.