What would the density of a sphere of radius 1.07 m have to be in order to produce a gravitational acceleration of 1.01 m/s2 at the surface of the sphere?
To determine the density of the sphere, we need to use the formula for gravitational acceleration:
g = (G * M) / r^2
where:
- g is the gravitational acceleration (1.01 m/s^2 in this case)
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the sphere
- r is the radius of the sphere (1.07 m in this case)
We also need to use the formula for the density of an object:
density = mass / volume
First, let's find the mass of the sphere:
Rearranging the formula for gravitational acceleration, we can solve for mass:
M = (g * r^2) / G
M = (1.01 m/s^2 * (1.07 m)^2) / (6.67430 × 10^-11 m^3 kg^-1 s^-2)
M ≈ 1.2397 * 10^9 kg
Next, we can find the volume of the sphere using the formula:
volume = (4/3) * π * r^3
volume = (4/3) * π * (1.07 m)^3
volume ≈ 5.1471 m^3
Finally, we can calculate the density using the mass and volume:
density = mass / volume
density = (1.2397 * 10^9 kg) / (5.1471 m^3)
density ≈ 2.4089 * 10^8 kg/m^3
Therefore, the density of the sphere would need to be approximately 2.4089 * 10^8 kg/m^3 in order to produce a gravitational acceleration of 1.01 m/s^2 at its surface.