Physics – 6.1.2 The Solar System | e-Consult

Answer:

1. Convert the orbital period to seconds: 12 hours * 60 minutes/hour * 60 seconds/minute = 43200 seconds

2. Use Kepler's Third Law of Planetary Motion, which relates orbital period (T), gravitational field strength (g), and orbital radius (r):

T² = (4π²/GM) * r³

Where:

• T = Orbital period (43200 s)
• G = Gravitational constant (6.67 x 10^-11 N m²/kg²)
• M = Mass of the planet (unknown, but we can rearrange the equation)
• r = Orbital radius (what we want to find)

3. Rearrange the equation to solve for r³:

r³ = (GM * T²) / (4π²)

4. Substitute the values:

r³ = (6.67 x 10^-11 N m²/kg² * M * (43200 s)²) / (4π²)

r³ = (6.67 x 10^-11 * M * 1.866 x 10⁹) / (4 * 9.87)

r³ = (1.24 x 10^-1 * M) / 39.48

r³ = 0.0315 * M

5. We are given the surface gravitational field strength (g = 9.8 m/s²) and the planet's mass (M = 2.0 x 10²⁷ kg). We can use the formula g = GM/r² to find r.

r² = GM/g

r² = (6.67 x 10^-11 * 2.0 x 10²⁷) / 9.8

r² = 1.36 x 10¹⁵ m²

r = √(1.36 x 10¹⁵) = 3.69 x 10⁷ m

Therefore, the radius of the spacecraft's orbit is approximately 3.69 x 10⁷ meters.