The International Space Station is 370 km above the Earth's surface (Re = 6378 km, Me = 5.98 x 10^24 kg). What is the linear velocity of the space station in m/s
mv²/(R+h) = GmM/(R+h)²
v=sqrt{GM/(R+h)} =….
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
To find the linear velocity of the space station, we can use the following formula:
v = √(GM/r)
Where:
- v is the linear velocity
- G is the gravitational constant (approximately 6.674 × 10^(-11) m^3/(kg·s^2))
- M is the mass of the Earth
- r is the distance between the space station and the center of the Earth
First, we need to find the value of r by adding the radius of the Earth (Re) with the altitude of the space station.
r = Re + h
Given that Re = 6378 km and the altitude of the space station is 370 km, we can convert these values into meters:
Re = 6378 km = 6378 × 1000 m = 6,378,000 m
h = 370 km = 370 × 1000 m = 370,000 m
Now, we can calculate r:
r = 6,378,000 m + 370,000 m = 6,748,000 m
Next, we can substitute the values of G, M, and r into the formula to find the linear velocity:
v = √((6.674 × 10^(-11) m^3/(kg·s^2)) × (5.98 × 10^24 kg) / 6,748,000 m)
Simplifying this expression, we have:
v = √(4.0029528251 × 10^(14) m^2/s^2)
Finally, we can calculate the square root to find the linear velocity:
v ≈ 6328.539 m/s
Therefore, the linear velocity of the International Space Station is approximately 6328.539 m/s.