The international space station ISS travels now with a speed of 7352 m/s. Calculate
a) the radius of its orbit;
b) the height of the orbit;
c) the period of its motion around Earth.
To calculate the radius of the International Space Station's orbit, we can use the fact that the centripetal force required to keep an object in circular motion is provided by the gravitational force between the object and Earth.
Let's start by calculating the radius of the orbit:
a) The centripetal force (F_c) is given by the equation F_c = (m * v^2) / r, where m is the mass of the ISS, v is the velocity of the ISS, and r is the radius of the orbit.
The gravitational force (F_g) between the ISS and Earth is given by the equation F_g = (G * M * m) / r^2, where G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of Earth, and m is the mass of the ISS.
Since the centripetal force is provided by the gravitational force, we can equate the two equations and solve for r:
F_c = F_g
(m * v^2) / r = (G * M * m) / r^2
Simplifying the equation, we can cancel out the mass of the ISS:
v^2 = (G * M) / r
Rearranging the equation to solve for r:
r = (G * M) / v^2
M, the mass of Earth, is approximately 5.972 × 10^24 kg.
Now, let's plug in the values:
r = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg) / (7352 m/s)^2
Calculating this expression will give us the radius of the orbit of the International Space Station.
b) To find the height of the orbit, we simply subtract the radius of the Earth (approximately 6,371 km) from the radius of the orbit calculated in part (a).
c) The period of its motion around Earth can be calculated using the formula T = 2πr / v, where T is the period and π is a constant approximately equal to 3.14159.
By substituting the values of r and v from part (a), we can calculate the period of the ISS's motion around Earth.