Find the distance (measured in earth radii, RE) from the center of the earth to a point outside the earth where the acceleration of gravity due to the earth is exactly one ninth of its value on the surface of the earth.
F=k/r^2 Since F is 1/9, it is apparent that r must be 3re
g(r) = constant *1/r^2
at r = 1 , g(1) = g
so
g(r) = g/r^2
g(r) = g/3^2 if r = 3
To find the distance from the center of the Earth to a point outside where the acceleration of gravity is one-ninth its surface value, we need to use the formula for gravitational acceleration.
The formula for gravitational acceleration is given by:
a = (G * M) / r^2,
where:
a is the acceleration due to gravity,
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 between the center of the Earth and the point outside.
We are given that the acceleration at this point (a') is one-ninth its surface value, so we can set up the following equation:
a' = (1/9) * (GM / r^2),
where a' is the acceleration we are given.
Now we can rearrange the equation to solve for r:
r^2 = (GM / (9 * a')),
Taking the square root of both sides gives us:
r = sqrt(GM / (9 * a')).
Substituting the known values into the equation, we get:
r = sqrt((6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2 * 5.972 × 10^24 kg) / (9 * a')).
Calculate the value of a', and then substitute it into the above equation to find the value of r. Finally, divide the value of r by the radius of the Earth (approximately 6,371 km) to get the value in Earth radii (RE).