How far (in meters) above the earth's surface will the acceleration of gravity be 27.0 % of what it is on the surface? can someone please just give me an equation to figure this out please I am lost
As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the expression g = GM/r^2 = µ/r^2 where GM = µ = the gravitational constant of the body (G = the Universal Gravitational Constant and M = the mass of the body) and r = the distance from the center of the body to the point in question.
With the earth's mean radius of 6378km and gravitational constant GM of 3.986365m^3/sec^2, the surface gravity g = GM/6378000^2 = 9.8m/sec^2. Therefore, the height above the earth's surface where g = .27(9.8) derives from r = sqrt[(GM/.27(9.8))/1000] - 6378.
(sqrt[1/.27]-1)(6.38*10^6)
Yes, I can help you with that. The equation you can use to solve this problem is the formula for gravitational acceleration diminishing with distance from the center of the Earth.
The equation is:
g' = g * (r / R)^2
where:
- g' is the acceleration of gravity at a given distance above the Earth's surface.
- g is the acceleration of gravity at the surface of the Earth (9.8 m/s^2).
- r is the distance above the Earth's surface (in meters) at which you want to find the acceleration of gravity.
- R is the radius of the Earth (6.371 × 10^6 m).
In this case, you need to find the distance (r) above the Earth's surface at which the acceleration of gravity (g') is 27.0% of what it is on the surface (g).
By substituting the given values into the equation and rearranging, you can solve for 'r':
0.27 = (r / R)^2
Take the square root of both sides to isolate 'r':
sqrt(0.27) = r / R
Now, calculate 'r' by multiplying the square root of 0.27 by the radius of the Earth (R):
r = sqrt(0.27) * R
To calculate how far above the Earth's surface the acceleration of gravity will be 27.0% of what it is at the surface, we can use the equation for gravitational acceleration:
g' = (G * M) / (r + h)^2
where:
g' is the gravitational acceleration at distance h above the surface,
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 radius of the Earth (approximately 6,371,000 meters),
h is the height above the Earth's surface.
We need to solve the equation for h when g' is 27.0% of the acceleration due to gravity at the surface.
Let's plug in the known values:
0.27 * g = (G * M) / (r + h)^2
Now, we need to rearrange the equation to solve for h:
(0.27 * g * (r + h)^2) = G * M
Expanding the square term:
0.27 * g * (r^2 + 2rh + h^2) = G * M
Distributing the 0.27 * g:
0.27 * g * r^2 + 0.27 * g * 2rh + 0.27 * g * h^2 = G * M
Rearranging to isolate the h terms:
0.27 * g * h^2 + 0.27 * g * 2rh + (0.27 * g * r^2 - G * M) = 0
Here, we obtain a quadratic equation in terms of h:
0.27 * g * h^2 + (0.27 * g * 2r)h + (0.27 * g * r^2 - G * M) = 0
To solve this equation, you can use the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / (2a)
Let's substitute the coefficients from our quadratic equation into the formula:
a = 0.27 * g
b = 0.27 * g * 2r
c = 0.27 * g * r^2 - G * M
Plug these values into the quadratic formula and solve for h. Remember to use the positive root, as we're looking for the positive distance above the Earth's surface.