At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 0.20%?
At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 2.0%?
At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 20%?
g = GM/r^2
so r = sqrt(GM/g)
g = 9.8 so .2% reduced = 9.8 - (.002*9.8)
G = 6.67e-11 and M = 5.98e24 for Earth.
Lather rinse repeat.
Oh sorry. And you need to subtract the radius of the earth from your answer to get how high above the surface.
What am I doing wrong...
for .2%
g=9.8-(.002*9.8)
.0196 =9.7804
r=sqrt(GM/g)
sqrt(3.9886*10^14)/9.7804
6386044.4
6386044.4-6371000=15044 m or aprx 15 km
?
To find the height above Earth's surface where the gravitational acceleration is reduced by a certain percentage, we can use the following formula:
h = (100 - p)R/g
Where:
- h is the height above Earth's surface
- p is the percentage decrease in gravitational acceleration
- R is the radius of the Earth (approximately 6,371 km)
- g is the gravitational acceleration at sea level (approximately 9.8 m/s^2)
Let's calculate the height for each percentage decrease:
1. For a decrease of 0.20% (p = 0.20):
h = (100 - 0.20) * 6371 km / 9.8 m/s^2 ≈ 322.97 km
2. For a decrease of 2.0% (p = 2.0):
h = (100 - 2.0) * 6371 km / 9.8 m/s^2 ≈ 635.20 km
3. For a decrease of 20% (p = 20):
h = (100 - 20) * 6371 km / 9.8 m/s^2 ≈ 32,627.55 km
So, the heights above Earth's surface where the gravitational acceleration is reduced by 0.20%, 2.0%, and 20% are approximately 323 km, 635 km, and 32,628 km, respectively.