What's the speed of the Hubble telescope when it is in orbit 598 km above the earths surface?
Did I not provide you with the means to calculate it? Or do you just want to answer mooch?
Oh no sorry didn't realize i repost but your formula was helpful
To determine the speed of the Hubble telescope when it is in orbit 598 km above the Earth's surface, we need to consider the concept of orbital velocity.
The orbital velocity of an object is the minimum velocity required to maintain a stable orbit around the Earth. It can be calculated using the following formula:
v = √(G * M / r)
Where:
v = orbital velocity
G = gravitational constant (approximately 6.67430 × 10^-11 m³kg⁻¹s⁻²)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
r = distance from the center of the Earth to the object's orbit
In this case, the distance from the center of the Earth to the Hubble telescope's orbit is the sum of the Earth's radius (approximately 6,371 km) and the altitude of the orbit (598 km). Therefore, r = 6,371 km + 598 km = 6,969 km = 6,969,000 meters.
Now we can substitute these values into the formula:
v = √(6.67430 × 10^-11 m³kg⁻¹s⁻² * 5.972 × 10^24 kg / 6,969,000 m)
After performing the calculation, the orbital velocity of the Hubble telescope in its orbit 598 km above the Earth's surface is approximately 7,679 meters per second (m/s).