The International Space Station (ISS) experiences an acceleration due to the Earth's gravity of 8.83 m/s2. What is the orbital period of the ISS?
So, I used 2pi r/v inputting 2 pi (6.37 x 10^6 + 380 x 10^3) / 529.6. 529.5 I got by muliplying 8.83 by 60 to hopefully translate to minutes. I got roughly 9065.9 and its telling me I'm wrong by a multiple of 10.
You almost had it, but there is a small mistake in your calculation method. Instead of directly converting the acceleration due to gravity into the orbital speed, we need to use the following formula for the orbital speed (v) first:
v^2 = GM/r, where G is the gravitational constant (6.674 * 10^-11 N m²/kg²), M is the mass of Earth (5.972 * 10^24 kg), r is the distance from the center of Earth to the ISS, which is the addition of the Earth's radius (6.37 * 10^6 m) and the ISS altitude (380 * 10^3 m).
r = 6.37 * 10^6 + 380 * 10^3 = 6.75 * 10^6 m
Now, we can find the orbital speed:
v^2 = (6.674 * 10^-11 N m²/kg²) * (5.972 * 10^24 kg) / (6.75 * 10^6 m)
v = √((6.674 * 10^-11 N m²/kg²) * (5.972 * 10^24 kg) / (6.75 * 10^6 m))
v ≈ 7,660 m/s
Now we can use the formula you mentioned for calculating the orbital period (T):
T = 2πr/v
T = 2π * (6.75 * 10^6 m) / (7,660 m/s)
T ≈ 5,548.8 sec or 92.48 min
So, the orbital period of the ISS is approximately 92.48 minutes.
To calculate the orbital period of the International Space Station (ISS), you can use the formula:
T = (2πr) / v
Where:
T is the orbital period
π is a mathematical constant approximately equal to 3.14159
r is the distance from the center of the Earth to the ISS's orbit (radius of the Earth plus altitude of the orbit)
v is the velocity of the ISS
First, let's calculate the value of r. The radius of the Earth is 6.37 x 10^6 meters. The altitude of the ISS orbit is given as 380 x 10^3 meters. Adding these values together, we get:
r = 6.37 x 10^6 + 380 x 10^3 = 6.75 x 10^6 meters.
Next, let's calculate the value of v. The acceleration due to the Earth's gravity is given as 8.83 m/s^2. However, this is the acceleration, not the velocity. To convert the acceleration to velocity, you need to multiply it by the time it takes to complete one revolution.
You mentioned multiplying the acceleration by 60 to get it in minutes, but you should actually divide it by 60 to get it in seconds. So, the value of v would be:
v = 8.83 m/s^2 / 60 = 0.1471 m/s.
Now, we can substitute the values of r and v into the formula to calculate the orbital period:
T = (2π × 6.75 x 10^6) / 0.1471.
Calculating this expression gives:
T = 9065.8612 seconds.
So, the orbital period of the International Space Station is approximately 9066 seconds.
To find the orbital period of the International Space Station (ISS), you can use the following formula:
T = 2π * √(r³ / GM)
Where:
T is the orbital period of the ISS,
π is a mathematical constant approximately equal to 3.14159,
r is the distance between the center of the Earth and the ISS (radius of the Earth plus the altitude of the ISS),
G is the gravitational constant, approximately equal to 6.67430 x 10^(-11)m³/(kg s²),
M is the mass of the Earth.
Let's calculate the orbital period step-by-step:
1. First, convert the altitude of the ISS from kilometers to meters:
380 x 10^3 meters = 380,000 meters.
2. Calculate the radius of the Earth plus the altitude of the ISS:
Radius of the Earth = 6.37 x 10^6 meters.
r = Radius of the Earth + altitude of the ISS
= 6.37 x 10^6 + 380,000
= 6.75 x 10^6 meters.
3. Convert the acceleration due to Earth's gravity from m/s² to m/s:
Acceleration due to Earth's gravity = 8.83 m/s².
Multiply by 60 to convert to m/minute:
Acceleration due to Earth's gravity = 8.83 m/s² * 60 s/min
= 529.8 m/min.
4. Calculate the orbital period by substituting the values into the formula:
T = 2π * √(r³ / GM)
= 2 * 3.14159 * √((6.75 x 10^6)³ / (6.67430 x 10^(-11) * (5.97 x 10^24)))
= 2 * 3.14159 * √(3.43 x 10^19 / 3.98 x 10^14)
= 2 * 3.14159 * √(861.8)
= 2 * 3.14159 * 29.37
= 184.65 minutes
So, the orbital period of the ISS is approximately 184.65 minutes.