Question

F. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? (1 day = 8.64x104s, ME = 6x1024kg, RE = 6.4x106m)

Answer 1

To calculate the height of the shuttle in a geosynchronous orbit, we can use Kepler's third law and the known values of Earth's mass, the universal gravitational constant, and the period of the orbit.

Step-by-step explanation:

To calculate the radius of a geosynchronous Earth satellite's orbit, we can use Kepler's third law. The period of the satellite's orbit is 1 day, which can be converted to seconds as 8.64 x 10₄ seconds. The formula we can use is: T₂ = rac{4π²r³}{GM}

By substituting the values for the period, Earth's mass (ME = 6 x 10²⁴ kg), and the universal gravitational constant (G = 6.67 x 10⁻¹¹ N²m/kg₃), we can solve for the radius (r) of the geosynchronous orbit.

Once the radius is calculated, we can subtract the radius of the Earth (RE = 6.4 x 10⁶ m) to find the height of the shuttle from the Earth's surface.

Answer 2

1.324 × 10⁷ m

Step-by-step explanation:

The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.

Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.

We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have

Rω² = GME/R²

R(2π/T)² = GME/R²

R³ = GME(T/2π)²

R = ∛(GME)(T/2π)²

RE + h = ∛(GMET²/4π²)

h = ∛(GMET²/4π²) - RE

substituting the values of the variables, we have

h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m

h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m

h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m

h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m

h = 19.64 × 10⁶ m - 6.4 × 10⁶ m

h = 13.24 × 10⁶ m

h = 1.324 × 10⁷ m