Question

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?

Answer 1

90400

Step-by-step explanation:

Answer 2

magnitude of the gravitational force is 9.04 ×
`10^(4)` N

Step-by-step explanation:

given data

altitude = A = 612 km = 612000 m

mass M = 11,100 kg

mass of the Earth m = 5.97 ×
`10^(24)` kg

Earth average radius = 6.38 ×
`10^(6)` m

to find out

magnitude of the gravitational force

solution

first we get here distance from space to centre of earth that is

distance = altitude + earth radius

distance = 612000 + 6.38 ×
`10^(6)` m

distance = 6.99 ×
`10^(6)` m

so now we get here magnitude of the gravitational force that is express as

magnitude of the gravitational force F =
`(G*M*m)/(distance^2)` ...........1

here G is gravitational constant that is 6.67 ×
`10^(-11)` Nm² /kg and M is mass of space and m is mass of earth

put here all value we get

F =
`(G*M*m)/(distance^2)`

F =
`(6.67*10^(-11)*5.97*10^(24)*11100)/((6.99*10^(6))^2)`

F = 9.04 ×
`10^(4)` N

so magnitude of the gravitational force is 9.04 ×
`10^(4)` N