What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/22 of its value at the Earth's surface?
_ m
1/22= (Re)^2/(R)^2
solve for R
whats re?
Re is the radius of the Earth-- about 6400 km
To find the distance from the Earth's center to a point outside the Earth where the gravitational acceleration is 1/22 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
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)
r is the distance from the center of the Earth.
At the Earth's surface, the gravitational acceleration can be denoted as g_s, and it is approximately 9.8 m/s^2.
We are given the gravitational acceleration at the point outside the Earth, which is 1/22 times its value at the Earth's surface. Therefore, we can write:
g_outside = (1/22) * g_s
Substituting the values, we get:
(1/22) * 9.8 m/s^2 = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg) / r^2
Simplifying the equation, we get:
r^2 = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg) / ((1/22) * 9.8 m/s^2)
Now, we can solve for r by taking the square root of both sides of the equation.