What is the distance in meters from the Earth's center to a point out-side the Earth where the gravitational acceleration due to the Earth is 1/20 of its value at the Earth's surface?

g at earths surface=GMe/re^2

1/20 g= GMe/r^2= g(1/(r/re)^2

(r/re )^2= 20
solve for r.

To find the distance from the Earth's center to a point outside the Earth where the gravitational acceleration is 1/20 of its value at the Earth's surface, we can use the following formula:

g = (G * M) / (r^2)

where:
g is the gravitational acceleration at a given point,
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg), and
r is the distance from the Earth's center to the point.

We know that at the Earth's surface, the gravitational acceleration is g_0 and is equal to approximately 9.8 m/s^2. Therefore, we can write the equation as:

g_0 = (G * M) / (R^2)

where R is the radius of the Earth.

Now, we want to find the distance r such that the gravitational acceleration is 1/20 (0.05) times the value at the Earth's surface. In other words, we want to solve the equation:

(1/20) * g_0 = (G * M) / (r^2)

Rearranging the equation, we have:

r^2 = (G * M) / ((1/20) * g_0)

Now, we can substitute the known values into the equation:

r^2 = (6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2 * 5.972 × 10^24 kg) / ((1/20) * 9.8 m/s^2)

Calculating the right side of the equation, we get:

r^2 = (6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2 * 5.972 × 10^24 kg) / (0.05 * 9.8 m/s^2)

Simplifying the expression, we find:

r^2 = 2.961840714 × 10^13 m^2

Taking the square root of both sides, we get:

r = √(2.961840714 × 10^13 m^2)

Evaluating the expression, the distance r is approximately equal to:

r ≈ 5.438350739 × 10^6 m

Therefore, the distance from the Earth's center to a point outside the Earth where the gravitational acceleration is 1/20 of its value at the Earth's surface is approximately 5.438 × 10^6 meters.