How much below the surface of the earth does the acceleration due to gravity become 1percent of its value at the earth surface? The earth is assume to be a sphere of radius 6.4×10^6m.
Beneath the surface, gravity is basically a linear function of the radius. So, at 1/100 the radius, g is reduced by a factor of 1/100
See the discussion here:
http://physics.stackexchange.com/questions/99117/why-gravity-decreases-as-we-go-down-into-the-earth
To determine at what depth below the surface of the Earth the acceleration due to gravity becomes 1 percent of its value at the Earth's surface, we can apply the concept of gravitational force and use the formula for gravitational acceleration:
g = G * (M / r^2)
where:
g = acceleration due to gravity
G = gravitational constant (approximately 6.67430 x 10^-11 N(m/kg)^2)
M = mass of the Earth
r = distance from the center of the Earth
We want to find the depth where g is 1 percent (0.01) of its value at the Earth's surface. Therefore, our equation becomes:
0.01 * g(surface) = G * (M / (r + d)^2)
where d represents the depth below the surface that we need to find.
We can simplify this equation by canceling out the common factors to solve for d:
0.01 * (G * (M / r^2)) = G * (M / (r + d)^2)
0.01 * (1 / r^2) = 1 / (r + d)^2
Next, we can cross multiply the equation:
0.01 * (r + d)^2 = r^2
Expand and rearrange the equation:
0.01 * (r^2 + 2rd + d^2) = r^2
0.01 * r^2 + 0.01 * 2rd + 0.01 * d^2 = r^2
0.01 * d^2 + 0.02 * rd + 0.99 * r^2 = r^2
0.01 * d^2 + 0.02 * rd = 0
Now, we solve for d:
0.01 * d^2 + 0.02 * rd = 0
d * (0.01 * d + 0.02 * r) = 0
Either d = 0 or 0.01 * d + 0.02 * r = 0
Since we are looking for a depth below the surface of the Earth, d cannot be zero. Therefore, we solve for the second equation:
0.01 * d + 0.02 * r = 0
0.01 * d = -0.02 * r
d = (-0.02 * r) / 0.01
Substitute the radius of the Earth given in the question (r = 6.4 × 10^6 m) into the equation:
d = (-0.02 * 6.4 × 10^6) / 0.01
d = (-0.128 × 10^6) / 0.01
d = -12.8 × 10^6 m = -1.28 × 10^7 m
Since we are looking for a depth, the negative value is not meaningful in this context. Therefore, we disregard the negative sign:
The depth below the surface of the Earth at which the acceleration due to gravity becomes 1 percent of its value at the Earth's surface is approximately 1.28 × 10^7 meters.