What is the magnitude of the gravitational force between the Earth and the Hubble Space Telescope (HST), which orbits the Earth? Assume the distance from the center of the Earth to the HST is 6,930 km, the mass of the Earth is 5.97 x 1024 kg, the mass of the HST is 1.11 x 104 kg, and G=6.67 x 10-11 Nm2/kg2.
As per newton's law of gravitation
F = GMm/(d^2)
Where G = 6.67 * 10^(-11) Nm^2kg^(-2) , d is the central distance
in metres and M and m are masses of two bodies
now, F = (6.67*10^(-11) * 1.11*10^4 * 5.97*10^24)/(6.93 * 10^6
)^2
(Radius of earth + altitude of telescope = central distance)
=92035.77519 N
hence, the earth exerts a pull of 92035.77519 N on the
telescope.
As per newton's law of gravitation
F = GMm/(d^2)
Where G = 6.67 * 10^(-11) Nm^2kg^(-2) , d is the central distance
in metres and M and m are masses of two bodies
now, F = (6.67*10^(-11) * 1110 * 5.97*10^24)/(6930000)^2
(Radius of earth + altitude of telescope = central distance)
=9203.5 N
What is the magnitude of the gravitational force between the Earth and the Hubble Space Telescope...
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