At what altitude above Earth's surface would the gravitational acceleration be 3.0 m/s2?
I will be happy to critique your thinking.
g = (G*M)/r²
Where G is The gravitational constant, M is the mass of earth, and r is the distance from the center of the earth.
Given this formula, you should be able to calculate the distance (for data about earth's mass and radius, search wikipedia for "Earth" and look in the table on the right)
To find 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 is the gravitational acceleration at a certain altitude,
G is the 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), and
r is the distance between the center of the Earth and the object.
We can rearrange the equation to solve for the distance r:
r = sqrt(G * M / g)
Now, substituting the given values, we have:
r = sqrt((6.67430 × 10^-11) * (5.972 × 10^24) / 3.0)
Calculating this equation will give us the distance from the center of the Earth to the required altitude above the surface. Subtracting the radius of the Earth (approximately 6,371 km) from this distance will give us the altitude above the Earth's surface.
Let's calculate it step by step:
1. Calculate G * M / g:
GM_over_g = (6.67430 × 10^-11) * (5.972 × 10^24) / 3.0
2. Calculate the square root of GM_over_g:
r = sqrt(GM_over_g)
3. Subtract the radius of the Earth from r to find the altitude above the surface.
Keep in mind that the radius of the Earth is given for reference, but the result may be slightly different due to the Earth not being a perfect sphere and varying gravitational acceleration at different altitudes.