At what altitude above the Earth's surface would your weight be two-thirds of what it is at the Earth's surface?

To solve this problem, we need to use the formula for gravitational force:

F = G * (m1 * m2) / r^2

where:

• F is the force of attraction between the two bodies,
• G is the gravitational constant,
• m1 and m2 are the masses of the two bodies,
• r is the distance between the centers of the two bodies.

On the surface of the Earth, your weight is the force of gravity acting on you. If we let m1 be your mass, m2 be the Earth's mass, and r be the Earth's radius, then your weight on the surface of the Earth is:

F_surface = G * (m1 * m2) / r^2

We want to find the altitude (h) above the Earth's surface where your weight is two-thirds of what it is on the surface. At this altitude, the distance between you and the Earth's center is r + h. So, the force of gravity at this altitude is:

F_altitude = G * (m1 * m2) / (r + h)^2

We know that F_altitude = 2/3 * F_surface, so we can set the two equations equal to each other and solve for h:

G * (m1 * m2) / (r + h)^2 = 2/3 * G * (m1 * m2) / r^2

Solving for h gives:

h = r * ( (1 / sqrt(2/3)) - 1)

The radius of the Earth (r) is approximately 6,371 kilometers. Plugging this into the equation gives:

h ≈ 6,371 km * ( (1 / sqrt(2/3)) - 1) ≈ 2,634 km

So, your weight would be two-thirds of what it is at the Earth's surface at an altitude of approximately 2,634 kilometers above the Earth's surface.