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

earth radius = 6.38*10^6

not 10^3

m g = G m M / r^2

so
g = G M / r^2

g' = G M /R^2

g'/g = 1/r'^2 / 1/r^2

g'/g = r^2/r'^2

1/20 = r^2/r'^2
so r'/r = sqrt 20 = 4.47
so r' = 4.47 * earth radius

Thanks for the help, but it said it wasn't right.... I did 4.47*(6.38*10^3)

Is that right?

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 formula for gravitational acceleration:

g = G * (M / r^2)

where g is the gravitational acceleration, G is the universal gravitational constant, M is the mass of the Earth, and r is the distance from the Earth's center.

First, let's denote the gravitational acceleration at the Earth's surface as g_surface, and the distance from the Earth's center to the point outside the Earth as r_point.

Given that the gravitational acceleration at the point outside the Earth is 1/20 of its value at the Earth's surface, we can write the following equation:

g_point = (1/20) * g_surface

We know that g_surface can be calculated by using the formula:

g_surface = G * (M / R^2)

where R is the radius of the Earth.

Substituting this equation into the previous one, we have:

(1/20) * g_surface = G * (M / r_point^2)

Now, we can solve for r_point:

r_point^2 = (M / (G * (1/20) * g_surface))

Simplifying the equation further, we get:

r_point^2 = (20 * M) / (G * g_surface)

Finally, we can take the square root of both sides to find r_point:

r_point = sqrt((20 * M) / (G * g_surface))

The mass of the Earth (M) and the universal gravitational constant (G) are known constants. The gravitational acceleration at the Earth's surface (g_surface) is approximately 9.8 m/s^2.

By plugging in these values, you can calculate the distance from the Earth's center to the desired point outside the Earth where the gravitational acceleration is 1/20 of its value at the Earth's surface.

28.5 *10^6 = 2.85 * 10^7 meters

this just isnt right...