On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 {\rm km} above the earth's surface, and we shall assume a circular orbit.
How many hours does it take this satellite to make one orbit? (T=?hours)
How fast (in {\rm km/s} )is the Aura spacecraft moving? (v=?km/s)
I've mistaken in the previous post.
New version:
m•a(norm) = GM/ [R+h] ^2,
m•v^2/(R+h) = GM/ [R+h] ^2,,
v = sqrt[GM/(R+h)] =sqrt [6.67•10^-11•5.97•10^24/7.1•10^6] =
= 7490 m/s =7.49 km/s
T = 2 •π(•R+h)/v =2 •π•7.1•10^6/7490= =5960 s =1.65 h
A boat propelled so as to travel with a speed of 0.50m/s in still water, moves directly ( in a straight line) across the river that is 60m wide. The rivers flows with a speed of 0.30m/s. How long in a seconds does it take the boat to across the river?
To find the time it takes for the satellite to make one orbit, we can use Kepler's third law, which relates the orbital period (T) to the radius of the orbit (r) using the formula:
T = 2π √(r³/GM)
Where:
T = Orbital period
r = Radius of the orbit
G = Gravitational constant
M = Mass of the Earth
In this case, the radius of the orbit is given as 705 km. The mass of the Earth is approximately 5.97×10^24 kg, and the gravitational constant is approximately 6.67430 x 10^-11 m^3⋅kg^−1⋅s^−2.
We need to convert the radius from km to m before plugging it into the formula. So, 705 km = 705,000 m.
Let's calculate the orbital period (T) in hours:
1. Convert the radius from km to meters:
r = 705,000 m
2. Plug the values into the formula:
T = 2π √((705,000)^3 / (6.67430 x 10^-11 * 5.97×10^24))
3. Solve for T:
T = 2π √(1.481 x 10^18 / 3.98742 x 10^14)
T = 2π √(3.714 x 10^3)
T ≈ 68.44 hours
Therefore, the Aura spacecraft takes approximately 68.44 hours to make one orbit.
Now, let's find the speed at which the spacecraft is moving (v) in km/s.
The speed of an object in circular orbit can be calculated using the formula:
v = 2πr / T
Using the orbit radius (r = 705,000 m) and the orbital period (T ≈ 68.44 hours = 246,384 seconds), we can calculate the speed (v).
1. Convert the radius from meters to km:
r = 705,000 m = 705 km
2. Convert the orbital period from hours to seconds:
T = 68.44 hours ≈ 246,384 seconds
3. Plug the values into the formula:
v = (2π * 705 km) / 246,384 seconds
4. Solve for v:
v = 9.018 km/s
Therefore, the Aura spacecraft is moving at a speed of approximately 9.018 km/s.
On the earth surface g = GM/R^2
And at the height h
g" = GM/ [R+h]^2
g" = R^2/ [ R+h]^2•g
g" = {R/ [ R+h] }^2•g
g" = {6378/ 7083}^2•9.81 = 7.95 m/s^2
T^2 = 4π^2•(R+h) /g"
T^2 = 4π^2•7083 /7.95
T = 187.54s = 0.052 hour.
v = 2•π •[R+h] / T
v = 2•π •[7.083] / 187.54s
v = 0.24 km /s