Question

The International Space Station orbits

the Earth (5.98 x 1024 kg) in a circle
of radius 6.77 x 106 m. What is its
orbital period in hours?
[?]hours
Enter

Answer 1

1.54 hours.

Step-by-step explanation:

To calculate the orbital period of the International Space Station (ISS) orbiting the Earth, we can use Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

First, let's calculate the gravitational parameter (μ) of the Earth, which is the product of the gravitational constant (G) and the mass of the Earth (M):

μ = G * M

where G is approximately 6.67430 x 10^-11 m^3/(kg*s^2) and M is the mass of the Earth, approximately 5.98 x 10^24 kg.

μ = (6.67430 x 10^-11 m^3/(kg*s^2)) * (5.98 x 10^24 kg)

Next, we can calculate the orbital period (T) using the formula:

T = 2π * √(a^3 / μ)

where a is the radius of the orbit, which is given as 6.77 x 10^6 m.

T = 2π * √((6.77 x 10^6 m)^3 / μ)

Now we can substitute the value of μ and calculate T:

T = 2π * √((6.77 x 10^6 m)^3 / ((6.67430 x 10^-11 m^3/(kg*s^2)) * (5.98 x 10^24 kg)))

Calculating this expression gives us the orbital period in seconds. To convert it to hours, we divide the result by 3600 seconds/hour:

T (in hours) = T (in seconds) / 3600

After performing the calculations, the orbital period of the International Space Station is approximately 1.54 hours.

Answer 2

the orbital period of an object in a circular orbit around the Earth with a radius of 6.77 x 10^6 meters is approximately 1.4159 hours.

Step-by-step explanation:

The semi-major axis of a circular orbit is equal to the radius of the orbit (r). In this case, the radius of the orbit is given as 6.77 x 10^6 meters.

Let's calculate the orbital period:

T^2 = (4π^2 / GM) * r^3

Where:

T = Orbital period

G = Gravitational constant (6.67430 x 10^-11 m^3 kg^-1 s^-2)

M = Mass of the Earth (5.98 x 10^24 kg)

r = Radius of the orbit (6.77 x 10^6 m)

Plugging in the values:

T^2 = (4π^2 / (6.67430 x 10^-11 * 5.98 x 10^24)) * (6.77 x 10^6)^3

Calculating this expression gives us:

T^2 ≈ 2.97576 x 10^13 seconds^2

Taking the square root of both sides to find T:

T ≈ √(2.97576 x 10^13) seconds

To convert this to hours, we divide by the number of seconds in an hour:

T ≈ √(2.97576 x 10^13) / 3600 hours

Calculating this expression gives us the orbital period:

T ≈ 1.4159 hours