What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/12 of its value at the Earth's surface?
Have u hrd of the inverse square law of gravity?
Multiply the distance from the center by sqrt12 (=3.464) and what do u get?
If you want the answer in km, multiply sqrt12 by the Earth's radius
To find the distance from the Earth's center to the point outside the Earth where the gravitational acceleration is 1/12 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 gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.
We want to find the distance, so we can rearrange the formula to solve for r:
r = √((G*M) / g)
To find the value of g at the Earth's surface, we can use the standard gravitational acceleration on Earth, which is approximately 9.8 m/s².
Substituting the values into the formula, we get:
r = √((G*M) / (1/12 * 9.8))
Now, we need to look up the values of G and M to calculate the distance:
- The gravitational constant, G, is approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2.
- The mass of the Earth, M, is approximately 5.972 x 10^24 kg.
Substituting these values, we can calculate the distance:
r = √((6.67430 x 10^-11 * 5.972 x 10^24) / (1/12 * 9.8))
Calculating this expression will give us the distance from the Earth's center to the point outside the Earth where the gravitational acceleration is 1/12 of its value at the Earth's surface.