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The space shuttle usually orbited Earth at an altitude of 320 km. Given the radius of the Earth (6.38x10⁶ m), the mass of the Earth (5.98x10²⁴ kg), the mass of the space shuttle (80,000 kg), and the mass of the Sun (1.98x10³⁰ kg), determine the time for one orbit of the shuttle about Earth. Provide your answer to two decimal places in hours.

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Final answer:

To calculate the time for one orbit of the space shuttle around Earth, Kepler's third law is used, modified for circular orbits. The sum of Earth's radius and the shuttle's altitude, the gravitational constant, and Earth's mass are factored into the equation. The calculated period is then converted from seconds to hours.

Step-by-step explanation:

To determine the time for one orbit of the space shuttle around Earth, we can use Kepler's third law, which implies that the square of the orbital period (T) is proportional to the cube of the semi-major axis of the orbit (r).

However, since we are dealing with a circular orbit close to the planet, we can simply apply the formula for circular orbital motion:

T = 2π ∗ sqrt[(r^3) / (G ∗ M)]

Where:

r is the sum of the earth radius (6.38x10⁶ m) and the altitude of shuttle (320x10³ m)

G is the gravitational constant (6.674x10⁻¹¹ m³/kg·s²)

M is the mass of the Earth (5.98x10²⁴ kg)

By plugging in the values, we get:

T = 2π ∗ sqrt[((6.38x10⁶ m + 320x10³ m)^3) / (6.674x10⁻¹¹ m³/kg·s² ∗ 5.98x10²⁴ kg)]

After computing the above expression, we convert the orbital period from seconds to hours. The result gives us the time it takes for one orbit of the shuttle about Earth, rounded to two decimal places.

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