The Hubble Space Telescope orbits Earth 621 km above Earth's surface. What is the period of the telescope's orbit? Give answer with 5 significant figures.

The period is given by

T = 2 * pi * (r^3/(G*M))^0.5

where T is the period, r is the radius, G is the gravitational constant, M is the mass of the earth

convert 621 km to meters
look up the Gravitational constant and the mass of the earth in kg

plug in the numbers, and solve

To calculate the period of an orbiting object, we can use the formula for the period of a satellite in circular orbit:

T = 2 * π * (R + h) / V

where:
T is the period of the orbit (in seconds)
π is a mathematical constant approximately equal to 3.14159
R is the radius of the Earth (approximately 6378 km)
h is the altitude of the object above the Earth's surface (in this case, 621 km)
V is the velocity of the satellite in its orbit

First, let's convert the altitude of the Hubble Space Telescope to kilometers:
h = 621 km

Next, let's find the total distance from the center of the Earth to the satellite:
R = 6378 km

To calculate the velocity of the satellite, we can use the formula:

V = √(G * M / (R + h))

where:
G is the universal gravitational constant (approximately 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2))
M is the mass of the Earth (approximately 5.972 × 10^24 kg)

To calculate the velocity, let's convert the units to meters:
G = 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)
M = 5.972 × 10^24 kg

Now, we can calculate the velocity of the satellite:
V = √((6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) * (5.972 × 10^24 kg) / (6378000 m))

Using a calculator, we find that V ≈ 7715.557 m/s.

Now we can calculate the period of the orbit:
T = 2 * π * (6378000 m + 621000 m) / 7715.557 m/s

Using a calculator, we find that T ≈ 5563.2737 seconds.

Therefore, the period of the Hubble Space Telescope's orbit is approximately 5563.27 seconds, rounded to 5 significant figures.

To calculate the period of an orbit, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit.

First, let's convert the altitude of the Hubble Space Telescope from kilometers to meters:
Altitude = 621 km = 621,000 meters

Next, we need to determine the semi-major axis of the orbit by adding the radius of the Earth to the altitude of the telescope:
Semi-major axis = Radius of Earth + Altitude

The radius of the Earth is approximately 6,371 km (6,371,000 meters). Adding this to the altitude:
Semi-major axis = 6,371,000 + 621,000 = 6,992,000 meters

Now we can calculate the period of the orbit using Kepler's Third Law equation:
Period² = (4π² / G) × (Semi-major axis)³

Where:
G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² (the gravitational constant)
π = 3.14159

Now we substitute the values into the equation:
Period² = (4π² / 6.67430 × 10⁻¹¹) × (6,992,000)³

Calculating this equation gives us:
Period² ≈ 8,556,016,295 seconds²

Finally, we take the square root of the period squared to get the period:
Period ≈ √(8,556,016,295) seconds

Rounding the result to 5 significant figures:
Period ≈ 92,550 seconds

Therefore, the period of the Hubble Space Telescope's orbit is approximately 92,550 seconds.