At what altitude above Earth's surface would the gravitational acceleration be 1.20 m/s2? (Take the Earth's radius as 6370 km.)
1.2 = GM/r^2
Solve for r and subtract out the earth's radius.
To find the altitude above Earth's surface where the gravitational acceleration is 1.20 m/s², we can use Newton's Law of Universal Gravitation.
The formula for gravitational acceleration (g) at a distance (r) from the center of the Earth can be given as:
g = G * M / r²
Where:
g = Gravitational acceleration
G = Universal gravitational constant (approximately 6.67430 × 10^-11 m^3/(kg*s^2))
M = Mass of the Earth (approximately 5.972 × 10^24 kg)
r = Distance from the center of the Earth (Earth's radius + altitude above the surface)
Given:
Earth's radius (R) = 6370 km
Now, let's solve for the altitude (h) above the Earth's surface where the gravitational acceleration is 1.20 m/s².
1. Convert the Earth's radius from kilometers to meters:
R = 6370 km = 6370 × 1000 m = 6,370,000 m
2. Substitute the values into the formula:
g = 1.20 m/s²
M = 5.972 × 10^24 kg
r = R + h
1.20 = (6.67430 × 10^-11 m^3/(kg*s^2)) * (5.972 × 10^24 kg) / (R + h)²
3. Rearrange the formula:
(R + h)² = ((6.67430 × 10^-11 m^3/(kg*s^2)) * (5.972 × 10^24 kg)) / 1.20
4. Simplify the right side of the equation:
(R + h)² = (4.00038 × 10^14 m^3/ s²)
5. Take the square root of both sides of the equation:
√(R + h)² = √(4.00038 × 10^14 m^3/ s²)
6. Solve for (R + h):
R + h = 2.00019 × 10^7 m
7. Subtract the Earth's radius, R, from both sides of the equation:
h = 2.00019 × 10^7 m - 6,370,000 m
h ≈ 13,630,000 meters
Therefore, the altitude above Earth's surface where the gravitational acceleration is approximately 1.20 m/s² is 13,630,000 meters.