Question

What is the attractive force between the earth and an 85kg piece of debris in an orbit of radius 27,500km with two rev/day?

Answer 1

Just applying gravitational force equation,

i.e., F = GmM/r²

where, m is mass of piece of debris , M is mass of the earth, r is the radius of orbit.

here, m = 85 kg , M = 6 × 10²⁴ Kg , r = 27,500km = 2.75 × 10^7 m and G = 6.67 × 10^-11 Nm²/Kg²

so, F = 6.67 × 10^-11 × 85 × 6 × 10²⁴/(2.75 × 10^7)²

= 6.67 × 510 × 10¹³/(2.75 × 2.75 × 10¹⁴)

= 449.8 × 10^-1 N

= 44.98 N ≈ 45 N

hence, attraction force between earth and piece of debris is 45N