What's the speed of the Hubble telescope when it is in orbit 598 km above the earths surface?
Its period is 96 min, so velcity must be...
velocity= 2PIradius/time=2PI(radEarth+598,000)/96
be certain to put radius Earth in meters to get velocity in m/min.
To find the speed of the Hubble telescope when it is in orbit 598 km above the Earth's surface, we need to use the principle of orbital mechanics. The speed of an object in orbit depends on the gravitational force acting on it and the radius of its orbit.
First, we need to calculate the radius of the orbit. The distance mentioned is the altitude above the Earth's surface, so to get the radius, we need to add the radius of the Earth (which is approximately 6,378 km) to the altitude.
Radius of orbit = Altitude + Radius of Earth
Radius of orbit = 598 km + 6,378 km
Radius of orbit = 6,976 km
Next, we can use the formula for orbital speed:
Orbital speed = √(G * Mass of the Earth / Radius of orbit)
where G is the gravitational constant, approximated as 6.67430 × 10^-11 m^3 kg^-1 s^-2, and the mass of the Earth is approximately 5.972 × 10^24 kg.
Converting the radius of the orbit from kilometers to meters:
Radius of orbit = 6,976 km * 1,000 = 6,976,000 m
Substituting the values into the formula:
Orbital speed = √(6.67430 × 10^-11 * 5.972 × 10^24 / 6,976,000)
Calculating the square root and multiplying the result by the appropriate units will give us the final answer.