The international space station (ISS) orbits Earth at an altitude of
4.08 x 10^5 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?
A) 7.66 x 10^3 m/s
B) 3.12 x 10^4 m/s
C) 7.91 x 10^3 m/s
D) 8.17 x 10^3 m/s
I don’t understand this at all 😭
To understand the velocity required for the International Space Station (ISS) to stay in its orbit, we need to consider the concept of centripetal force and gravitational force.
The centripetal force is responsible for keeping the ISS in orbit, and it is provided by the gravitational force between the Earth and the ISS. This force is given by the equation:
F = (G * m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the Earth and ISS respectively, and r is the distance between the centers of the Earth and ISS.
In this case, we can assume the mass of the ISS (m2) to be negligible compared to the mass of the Earth (m1), so the equation simplifies to:
F = (G * m1) / r^2
Now, the centripetal force required to keep the ISS in orbit is given by:
Fc = (m2 * v^2) / r
where Fc is the centripetal force, m2 is the mass of the ISS, v is the velocity of the ISS, and r is the distance between the Earth's surface and the ISS.
Since the centripetal force is provided entirely by the gravitational force, we can equate these two forces:
(G * m1) / r^2 = (m2 * v^2) / r
We can rearrange the equation to solve for v:
v^2 = (G * m1) / r
Taking the square root of both sides gives:
v = sqrt((G * m1) / r)
Now we can substitute the known values into the equation:
m1 = mass of the Earth = 5.972 × 10^24 kg
r = distance above the Earth's surface = 4.08 × 10^5 m
G = gravitational constant = 6.67430 × 10^-11 m^3 kg^-1 s^-2
v = sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg) / (4.08 × 10^5 m))
Calculating this, the velocity required for the ISS to stay in its orbit is approximately 7.66 × 10^3 m/s. This matches option A) 7.66 × 10^3 m/s.