The Hubble Space Telescope orbits the Earth at an approximate altitude of 612 km. Its mass is 11,100 kg and the mass of the Earth is 5.97×1024 kg. The Earth's average radius is 6.38×106 m. What is the magnitude of the gravitational force that the Earth exerts on the Hubble?
To calculate the magnitude of the gravitational force that the Earth exerts on the Hubble Space Telescope, you will need to use Newton's law of universal gravitation. The formula is as follows:
F = (G * m1 * m2) / r^2
Where:
F - Magnitude of the gravitational force
G - Gravitational constant (6.67430 × 10^-11 N m^2 kg^-2)
m1 - Mass of the first object (Earth)
m2 - Mass of the second object (Hubble Space Telescope)
r - Distance between the centers of the two objects
Given:
Mass of the Earth (m1) = 5.97 × 10^24 kg
Mass of the Hubble Space Telescope (m2) = 11,100 kg
Altitude of the Hubble Space Telescope (r) = 612 km = 612,000 m (since the altitude is given above the Earth's surface, we need to add the Earth's radius to get the actual distance)
Now, let's calculate the magnitude of the gravitational force:
r = Earth's average radius + Altitude of the Hubble Space Telescope
= 6.38 × 10^6 m + 612,000 m
= 6.992 × 10^6 m
F = (6.67430 × 10^-11 N m^2 kg^-2 * 5.97 × 10^24 kg * 11,100 kg) / (6.992 × 10^6 m)^2
Simplifying the equation:
F = (5.97 × 10^24 kg * 11,100 kg) / (6.992 × 10^6 m)^2
Calculating the numerator:
Numerator = 5.97 × 10^24 kg * 11,100 kg
= 6.633 × 10^28 kg^2
Calculating the denominator:
Denominator = (6.992 × 10^6 m)^2
= 4.885 × 10^13 m^2
Plugging the values back into the equation:
F = 6.633 × 10^28 kg^2 / 4.885 × 10^13 m^2
Simplifying further and expressing it in scientific notation:
F = 1.358 × 10^15 N
Therefore, the magnitude of the gravitational force that the Earth exerts on the Hubble Space Telescope is approximately 1.358 × 10^15 Newtons.