Find the altitudes above the Earth's surface where Earth's gravitational field strength would be (a) five-sixths (enter first) and (b) one-sixth of its value at the surface.

The gravitational field strength (g) at a distance (r) from the center of the Earth is given by the formula:

g = GM/r^2

where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth. At the surface of the Earth, r is the radius of the Earth (R), so the gravitational field strength at the surface (g0) is:

g0 = GM/R^2

(a) If we want to find the altitude (h) where the gravitational field strength is five-sixths of its value at the surface, we set g = 5g0/6 and solve for r:

5g0/6 = GM/r^2

Substituting g0 = GM/R^2 gives:

5/6 = (R/r)^2

Solving for r gives:

r = R*sqrt(6/5)

The altitude h is the distance above the Earth's surface, so:

h = r - R = Rsqrt(6/5) - R = R(sqrt(6/5) - 1)

(b) If we want to find the altitude where the gravitational field strength is one-sixth of its value at the surface, we set g = g0/6 and solve for r:

g0/6 = GM/r^2

Substituting g0 = GM/R^2 gives:

1/6 = (R/r)^2

Solving for r gives:

r = R*sqrt(6)

The altitude h is the distance above the Earth's surface, so:

h = r - R = Rsqrt(6) - R = R(sqrt(6) - 1)

Note: This is a simplified model that assumes the Earth is a perfect sphere and that its mass is uniformly distributed. In reality, the Earth is not a perfect sphere and its mass is not uniformly distributed, so the actual altitudes would be slightly different.