At what altitude above Earth's surface would the gravitational acceleration be 3.0 m/s2?
I am not going to do all this arithmetic
F/m = g apparent = 3 = G Mearth/(Rearth +h)^2
where we know
9.8 = G Mearth/Rearth^2
so
9.8/3 = (Rearth+h)^2 / Rearth^2
Let:
m = a small mass
M = mass of the earth
r = distance from the center of mass m to the earth's center.
Constants:
G = 6.67x10^-11 m^3/kg.s^2.
M = 5.98x10^24 kg
R = 6.38x10^6 m (Earth's radius)
Law Of Gravitation:
F = GmM/r^2
F = m(GM/r^2) = mg
GM/r^2 = 3.00 m/s^2
Solve for r and subtract the radius of the earth.
To determine the altitude above Earth's surface where the gravitational acceleration is 3.0 m/s^2, we can use the formula for gravitational acceleration:
g = (G * M) / (R^2)
Where:
g = gravitational acceleration (in m/s^2)
G = gravitational constant (approximately 6.67 x 10^-11 N*m^2/kg^2)
M = mass of the Earth (approximately 5.97 x 10^24 kg)
R = distance from the center of the Earth
We need to solve for R, the distance from the center of the Earth where the gravitational acceleration is 3.0 m/s^2.
R^2 = (G * M) / g
R^2 = (6.67 * 10^-11 N*m^2/kg^2 * 5.97 * 10^24 kg) / (3.0 m/s^2)
R^2 = (3.98 * 10^14 N*m^2) / (3.0 m/s^2)
R^2 = 1.33 * 10^14 m^2
R ≈ sqrt(1.33 * 10^14) m
R ≈ 3.65 * 10^7 m
Therefore, the altitude above Earth's surface where the gravitational acceleration is 3.0 m/s^2 is approximately 3.65 * 10^7 meters.
To determine the altitude above Earth's surface at which the gravitational acceleration would be 3.0 m/s^2, we can use the formula for gravitational acceleration:
g = G * (M / r^2)
Where:
- g is the gravitational acceleration
- G is the universal gravitational constant (approximately 6.67430 × 10^-11 m^3/(kg s^2))
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- r is the distance from the center of the Earth to the altitude of interest
We can rearrange the formula to solve for r:
r = √(G * M / g)
Substituting the given values:
r = √((6.67430 × 10^-11 m^3/(kg s^2)) * (5.972 × 10^24 kg) / (3.0 m/s^2))
Calculating the expression:
r ≈ 6.37 × 10^6 meters
Therefore, at an altitude of approximately 6.37 × 10^6 meters above Earth's surface, the gravitational acceleration would be 3.0 m/s^2.