Final answer:
The weight of the Hubble Space Telescope in its orbit can be calculated using Newton's law of universal gravitation. By substituting the given values into the gravitational force formula, we find that the telescope's weight is approximately 9.82 x 10^4 Newtons at an altitude of 598 km above the Earth's surface.
Step-by-step explanation:
To calculate the weight of the Hubble Space Telescope when it is in its orbit above Earth's surface, we will use Newton's law of universal gravitation. The weight of an object in orbit is essentially the gravitational force acting on it. This force can be calculated using the formula:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force (weight of the telescope in orbit)
G is the gravitational constant (6.674x10^-11 N(m^2)/(kg^2))
m1 is the mass of the Earth (5.98×10^24 kg)
m2 is the mass of the Hubble Space Telescope (11,600 kg)
r is the distance from the center of the Earth to the telescope, which includes Earth's radius plus the altitude of the telescope in orbit (6.37×10^6 m + 598×10^3 m)
Calculating the distance r:
r = Earth's radius + telescope's altitude above Earth's surface
r = 6.37×10^6 m + 598×10^3 m
r = 6.968×10^6 m
Now, we can substitute all the values into the formula:
F = (6.674×10^-11 N(m^2)/(kg^2)) * ((5.98×10^24 kg) * (11,600 kg)) / (6.968×10^6 m)^2
After calculating, we find:
F ≈ 9.82 x 10^4 N
This is the weight of the Hubble Space Telescope at the given altitude above Earth's surface. It is important to remember that while the telescope has weight due to the gravitational pull, it is in freefall around the Earth, which is why it does not crash into the Earth's surface, thereby giving the experience of weightlessness.