A satellite orbits at an average altitude of h = 409 km. What is its orbital speed? Use REarth = 6370 km and
mEarth = 5.98 x 10^24 kg
set centripetal force = gravitational force
GMm/(r+h)^2=mv^2/(r+h)
v= 2PI (r+h)/24hrsinSeconds
change all kilometers to meters. Watch units.
can you please explain the v=...part i don't understand what you mean by 24hrsinseconds?
is GMm gravity (9.81) multiplied by the mEarth?
To calculate the orbital speed of a satellite, we can use the following formula:
V = sqrt((G * mEarth) / (REarth + h))
where:
V is the orbital speed,
G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2),
mEarth is the mass of the Earth (approximately 5.98 x 10^24 kg),
REarth is the radius of the Earth (approximately 6370 km),
h is the altitude of the satellite above the Earth's surface.
Let's substitute the values into the formula:
V = sqrt((6.67 x 10^-11 N m^2/kg^2 * 5.98 x 10^24 kg) / (6370 km + 409 km))
First, let's convert the radius of the Earth to meters:
REarth = 6370 km = 6370 x 1000 m = 6,370,000 m
Next, let's convert the altitude to meters:
h = 409 km = 409 x 1000 m = 409,000 m
Now we can substitute the values:
V = sqrt((6.67 x 10^-11 N m^2/kg^2 * 5.98 x 10^24 kg) / (6,370,000 m + 409,000 m))
V = sqrt((3.9826 x 10^14 N m^2/kg) / 6,779,000 m)
V = sqrt(5.8752 x 10^7 m^2/s^2)
V ≈ 7660 m/s
Therefore, the orbital speed of the satellite is approximately 7660 m/s.