The International Space Station, launched in November 1998, makes 15.65 revolutions around the earth each day in a circular orbit.
Apply Newton's second law to the space station and find its height (in kilometers) above the earth's surface.
T = 2(Pi)sqrt[r^3/µ)
T = orbit period in seconds
r = radial distance of station from Earth's center
µ = Earth's gravitational constant = 3.986365x10^14m^3/sec.^2
Solve for r
To find the height of the International Space Station above the Earth's surface using Newton's second law, we need to consider the gravitational force acting on the spacecraft.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force acting on the space station is the gravitational force.
The gravitational force between two objects can be given by the equation:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force
G is the gravitational constant (6.674 × 10^-11 N m^2/kg^2)
m1 is the mass of the space station
m2 is the mass of the Earth
r is the distance between the center of the two objects
In this case, we assume that the mass of the space station is much smaller than the mass of the Earth, so we can neglect its effect on the Earth's motion. Thus, the force acting on the space station is solely due to the Earth's gravity.
Considering that the space station is in a circular orbit, which means it experiences centripetal acceleration, we can equate the gravitational force to the centripetal force:
F = m * a
Where:
m is the mass of the space station
a is the centripetal acceleration
The centripetal acceleration can be calculated using the formula:
a = v^2 / r
Where:
v is the orbital velocity of the space station
r is the distance between the center of the Earth and the space station (height above the Earth's surface)
Since the space station makes 15.65 revolutions around the Earth per day, we can calculate its orbital period and thus its orbital velocity:
Orbital period = 24 hours / 15.65 revolutions
= 1.534 hours per revolution
= 5514.24 seconds per revolution
The orbital velocity can be calculated using the formula:
v = 2πr / T
Where:
π is approximately 3.14159
r is the radius of the Earth plus the height of the space station
T is the orbital period
Using these equations, we can solve for the height of the space station above the Earth's surface.
To find the height of the International Space Station (ISS) above the Earth's surface using Newton's second law, we need to consider the centripetal force acting on the ISS.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
In this case, the net force is the centripetal force, which is given by the formula:
F = (m * v^2) / r
where m is the mass of the ISS, v is its velocity, and r is its radius of orbit (which is the distance from the Earth's center to the ISS).
The velocity (v) of the ISS can be calculated by dividing the distance it travels in one revolution by the time it takes to complete one revolution.
The time it takes to complete one revolution can be calculated by dividing the total number of minutes in a day (24 * 60 = 1440) by the number of revolutions the ISS makes in a day.
Hence,
t = 1440 / 15.65
Now, we can calculate the velocity (v) of the ISS:
v = (2 * π * r) / t
where π is a mathematical constant (approximately 3.14159).
Given that the radius of orbit (r) is the sum of the Earth's radius and the height of the ISS above the Earth's surface, we can express it as:
r = R + h
where R is the mean radius of the Earth (approximately 6371 km) and h is the height of the ISS above the Earth's surface.
Finally, we can substitute the values into the centripetal force formula:
F = (m * v^2) / (R + h)
Since the ISS is in a state of weightlessness, the net force acting on it is zero:
F = 0
Now, we can solve for the height (h):
0 = (m * v^2) / (R + h)
Rearranging the equation:
h = -R + (m * v^2) / F
Substituting the appropriate values and solving the equation will yield the height of the ISS above the Earth's surface.