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%?
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
6386044.4-6370000=16044 m or aprx 16 km
am i not using the right equation?
g'=g*(re/(re+r))^2
so if you want g' to be say 1percetn less, then g'=.99g
or .99=(re/(re+r))^2
re/(re+r)= sqrt .99
then solve for r, which is the height above the Earth's surface
Your logic is perfectly sound. The problem is that using rounded figures like 5.98e24 for mass and 6.34e6 for radius gets you an initial value for earth's gravity as 9.83. Using less rounded figures like mass as 5.972kg gets closer to the truth (9.81).
But again we're dealing with really big numbers so getting .2% from these big numbers is subject to error. Using a little more exact figures your way gets a figure of about 11800m. But your way of doing it is completely correct.
Your approach is correct for finding the height above Earth's surface where the gravitational acceleration is reduced by a certain percentage. However, there is a mistake in the equation you used to find 'r', the distance from the center of the Earth to the point above the surface.
To find the correct equation, we need to use the formula for the acceleration due to gravity at a given height:
g = GM/(r^2)
where g is the gravitational acceleration at the height, M is the mass of the Earth, and r is the distance from the center of the Earth to the point above the surface.
To solve for 'r', we can rearrange the equation as:
r = sqrt(GM/g)
Let's go through each case:
1. For a reduction of 0.20%:
g = 9.8 * (1 - 0.002)
g = 9.8 * 0.998 = 9.7804 m/s^2
r = sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / 9.7804)
r ≈ 6,371,035 meters or approximately 6,371 kilometers
So, at a height of approximately 6,371 kilometers above Earth's surface, the gravitational acceleration is reduced by 0.20%.
2. For a reduction of 2.0%:
g = 9.8 * (1 - 0.02)
g = 9.8 * 0.98 = 9.604 m/s^2
r = sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg ) / 9.604)
r ≈ 6,444,733 meters or approximately 6,445 kilometers
At a height of approximately 6,445 kilometers above Earth's surface, the gravitational acceleration is reduced by 2.0%.
3. For a reduction of 20%:
g = 9.8 * (1 - 0.20)
g = 9.8 * 0.80 = 7.84 m/s^2
r = sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / 7.84)
r ≈ 8,545,856 meters or approximately 8,546 kilometers
At a height of approximately 8,546 kilometers above Earth's surface, the gravitational acceleration is reduced by 20%.
Please note that these calculations assume a uniform spherical Earth and neglect other gravitational influences like the Moon and the Sun.