What is the magnitude of g at a height above Earth's surface where free fall acceleration equals 6.5m/s^2?
Well, if the free fall acceleration is 6.5 m/s^2, it means gravity is having a blast but taking it easy. It's like a roller coaster that's a bit sluggish, not too fast, not too furious. So, at that height above Earth's surface, the magnitude of g would be 6.5 m/s^2.
To calculate the magnitude of the acceleration due to gravity, we can use the formula:
g = G * (M / r^2)
where:
g is the magnitude of the acceleration due to gravity,
G is the gravitational constant (approximately 6.67430 × 10^-11 N m^2 / kg^2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg), and
r is the distance between the center of the Earth and the height above the surface.
In this case, we know that the free fall acceleration at this height is 6.5 m/s^2. This means that the net force acting on an object in free fall at this height is equal to its mass multiplied by this acceleration.
We can set up the following equation:
6.5 m/s^2 = G * (M / r^2)
We can rearrange this equation to solve for r:
r^2 = (G * M) / 6.5 m/s^2
r = sqrt((G * M) / 6.5 m/s^2)
Now, substituting the known values, we can calculate the magnitude of g at this height:
G = 6.67430 × 10^-11 N m^2 / kg^2
M = 5.972 × 10^24 kg
6.5 m/s^2 = 6.67430 × 10^-11 N m^2 / kg^2 * (5.972 × 10^24 kg / r^2)
Now, solving for r:
r^2 = (6.67430 × 10^-11 N m^2 / kg^2 * 5.972 × 10^24 kg) / 6.5 m/s^2
r^2 = 4.449 × 10^14 m^2
Taking the square root:
r = sqrt(4.449 × 10^14 m^2)
r ≈ 6.66 × 10^7 meters
Therefore, the magnitude of g at a height above Earth's surface where the free fall acceleration equals 6.5m/s^2 is approximately 6.66 × 10^7 meters.
To find the magnitude of acceleration due to gravity, g, at a certain height above the Earth's surface, we can use the formula:
g' = g * (R / (R + h))^2
Where:
g' is the acceleration due to gravity at the specified height
g is the acceleration due to gravity at the Earth's surface (approximately 9.8 m/s^2)
R is the radius of the Earth (approximately 6,371 kilometers or 6,371,000 meters)
h is the height above the Earth's surface
In this case, we are given that the free fall acceleration at the specified height is 6.5 m/s^2. We need to find the corresponding value of g.
Rearranging the formula, we can solve for g:
g = g' * ((R + h) / R)^2
Substituting the given values:
g = 6.5 * ((6,371,000 + h) / 6,371,000)^2
Now, we can calculate the value of g by plugging in the height value.
Assuming you are seeking the altitude where gravity = g = 6.5m/s^2:
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.
Therefore, r = sqrt[µ/g]
= sqrt[3.986365x10^14/6.5
= sqrt[6.132869x10^13] = 7,831,263m
= 7831.263km.
The distance above the surface is therefore h = ~(7831 - 6378) = 1453km.