At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 0.60%?
less than one percent? Are you certain you wnat that fraction? .6%=,006
1-.006=.994=(re/(re+h))^2 solve for h.
I still don't get it. cna you explain
To find the height above Earth's surface at which the gravitational acceleration is reduced by 0.60%, we need to use the formula for gravitational acceleration:
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 Earth (approximately 5.972 × 10^24 kg), and
r is the distance from the center of Earth to the object.
We are given that the gravitational acceleration at this height is reduced by 0.60%. Therefore, we can write the equation as:
(1 - 0.006)g = G * (M / (r + h)^2)
where h is the height above Earth's surface that we need to find.
Simplifying the equation, we get:
g = G * (M / (r + h)^2)
Dividing the two equations, we obtain:
(1 - 0.006)g / g = G * (M / (r + h)^2) / G * (M / r^2)
Simplifying further, we have:
(1 - 0.006) / 1 = (r^2 / (r + h)^2)
Now, let's solve for h:
(1 - 0.006) / 1 = (r^2 / (r + h)^2)
0.994 = (r^2 / (r + h)^2)
Taking the square root of both sides:
sqrt(0.994) = r / (r + h)
Squaring both sides:
0.994 = (r / (r + h))^2
Taking the reciprocal:
1 / 0.994 = (r + h) / r
Simplifying:
1.006 = (r + h) / r
Cross-multiplying:
r + h = 1.006r
Subtracting r from both sides:
h = 0.006r
Now, substituting the known value of the radius of Earth (r ≈ 6.371 × 10^6 m):
h ≈ 0.006 * 6.371 × 10^6 m
Calculating:
h ≈ 3.8226 × 10^4 m
Therefore, at a height of approximately 38,226 meters (or 38.23 km) above Earth's surface, the gravitational acceleration is reduced by 0.60%.