at what altitude above the earth's surface is the free fall acceleration equal to three quarters of its value at the surface ?
g wherever = some constant k/r^2
let r = earth radius
let R = radius with 3/4 g
then we want R-r
g surface = k /r^2
g up = k/R^2
g up/g surface = k/R^2/k/r^2 = r^2/R^2 = 3/4
so
R/r = sqrt 4 /sqrt 3
look up r, radius of earth and you can find R, then subtract.
To find the altitude at which the free fall acceleration is equal to three quarters (3/4) of its value at the Earth's surface, we need to consider the relationship between acceleration due to gravity and altitude.
The acceleration due to gravity can be approximated using the formula:
a = g * (R / (R + h))^2
where:
a is the acceleration at altitude h,
g is the acceleration due to gravity at the Earth's surface (9.8 m/s^2),
R is the Earth's radius (6,371 km), and
h is the altitude from the Earth's surface.
We need to find the altitude h at which the expression for acceleration becomes 3/4 of g. So, we can set up the following equation:
3/4 * g = g * (R / (R + h))^2
Now, we can solve this equation to find the value of h.
1. Start by canceling out g:
3/4 = (R / (R + h))^2
2. Take the square root of both sides:
√(3/4) = R / (R + h)
3. Square both sides to remove the square root:
3/4 = [R / (R + h)]^2
4. Cross multiply:
3(R + h)^2 = 4R^2
5. Expand and simplify:
3(R^2 + 2Rh + h^2) = 4R^2
6. Distribute and rearrange:
3R^2 + 6Rh + 3h^2 = 4R^2
7. Subtract 4R^2 from both sides:
3h^2 + 6Rh - R^2 = 0
Now we have a quadratic equation. We can solve it using the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / 2a
In this case, the equation is in the form of ah^2 + bh + c = 0, so:
a = 3
b = 6R
c = -R^2
Plugging the values into the quadratic formula, we get:
h = [-(6R) ± √((6R)^2 - 4(3)(-R^2))] / (2 * 3)
Simplifying further:
h = [-6R ± √(36R^2 + 12R^2)] / 6
h = [-6R ± √(48R^2)] / 6
h = [-6R ± 4R√3] / 6
h = R (-1 ± (2√3)/3)
Now we have two possible solutions for altitude, one with a positive sign and another with a negative sign. Since we're looking for a positive altitude, we choose the positive sign:
h = R * (-1 + (2√3)/3)
Using the approximate value for the Earth's radius (R = 6,371 km), we can substitute it into the equation to find the altitude:
h ≈ 6,371 km * (-1 + (2√3)/3)
Calculating the numerical value,
h ≈ -6,371 km + (2 * 6,371 km * √3)/3
h ≈ -6,371 km + (12,742 km * √3)/3
h ≈ -6,371 km + (4,430 km * √3)
h ≈ -6,371 km + 7,635 km
h ≈ 1,315 km
Therefore, at an altitude of approximately 1,315 kilometers above the Earth's surface, the free fall acceleration is equal to three quarters (3/4) of its value at the surface.