Calculate the speed needed by a spacecraft to maintain a circular orbit at about 350 km above Earth’s surface.
(it's a project where we create our own spaceship. let's say the craft is 500kg)
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Not sure about the formula to find this. All I know is that it may have something to do with the escape velocity of earth, which is about 1.1x10^4m/s.
mass * centripetal acceleration = mass * gravity at that height
You might as well assume g = 9.8 I suppose because 350 km is pretty small compared to earth radius.
In any case you see the mass does not matter. A feather would stay in orbit beside you.
m v^2/r = m g
v^2 = gr
r = 350,000 + radius of earth
where earth radius = 6,380,000 meters
to be more accurate use
g at height = G Me/r^2 = 6.67*10^-11 * 5.98*10^24/r^2
The orbital speed for low Earth orbit is 71% of the escape velocity. You are right; there is a connection.
You can derive it using the formula:
V^2/R = GM/R^2
V = sqrt(GM/R)
Where R is the distance from the center of the Earth, which in this case is 6371 + 350 = 6721 km. G is the universal constant of gravity,
6.673*10−11 N m^2 kg^−2, and M is the mass of the Earth, 5.98 × 10^24 kg.
V = 7.70*10^3 m/s
To calculate the speed needed by a spacecraft to maintain a circular orbit, you can use the formula for orbital velocity. The orbital velocity of an object in a circular orbit can be calculated using the following formula:
V = sqrt(G * M / R)
Where:
- V is the orbital velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the central body (in this case, the Earth)
- R is the distance between the spacecraft and the center of the Earth
In this case, the spacecraft is at a distance of 350 km above the Earth's surface, which we can convert to meters (1 km = 1000 m).
R = 350,000 m + radius of the Earth
To find the radius of the Earth, it is approximately 6,371 km.
R = 350,000 m + 6,371,000 m
Now, we can plug in the values into the formula to calculate the orbital velocity:
V = sqrt(G * M / R)
= sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / (350,000 m + 6,371,000 m))
After performing the calculations, we find the orbital velocity needed to maintain a circular orbit at about 350 km above Earth's surface is approximately 7,667 m/s.
To calculate the speed needed by a spacecraft to maintain a circular orbit around Earth at a specific altitude, you can use the following formula:
v = √(GM/r)
Where:
- v represents the required orbital speed
- G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
- M represents the mass of Earth (approximately 5.972 × 10^24 kg)
- r is the radius of the orbit (which is the sum of Earth's radius and the altitude of the orbit)
In this case, you want to calculate the speed for an altitude of 350 km, so the radius (r) would be the sum of the Earth's radius (6,371 km) and the altitude (350 km).
1. Convert the radius (r) into meters:
r = (Earth's radius + altitude) in meters = (6,371 km + 350 km) × 1000 = 6,721,000 meters
2. Now we can plug the values into the formula:
v = √((6.67430 × 10^-11 m^3 kg^-1 s^-2) × (5.972 × 10^24 kg) / (6,721,000 m))
Simplifying the equation:
v = √(3.986004418 x 10^14 m^3 s^-2 kg^-1 / 6,721,000 m)
Taking the square root of the fraction:
v ≈ √59,391.1928 m^2 s^-2 ≈ 243.67 m/s (rounded to two decimal places)
Therefore, the speed needed by a spacecraft to maintain a circular orbit at an altitude of 350 km above Earth's surface is approximately 243.67 m/s.
Please note that in reality, this calculation assumes a perfect circular orbit and neglects other factors such as atmospheric drag, other celestial bodies' gravitational influence, and the spacecraft's thrust and propulsion systems.