Since the spacecraft is two earth radii the surface of the earth, it is three earth radii above the center.
Given: Radius of the earth re = 6.38 x 10⁶ m r = 1.91 x 10⁷ m
Mass of the spacecraft Ms = 1600 Kg
Mass of the earth Me = 5.98 x 10²⁴ Kg
G = 6.67 X 10⁻¹¹ N.m²/Kg²
Formula: F = GMeMs/r²
F = (6.67 X 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)(1,600 Kg)/(1.91 x 10⁷ m)²
F = 6.38 X 10¹⁷ N/3.65 X 10¹⁴ m²
F = 1,748.5 N
Answer:
= 1.7 Joules
Explanation:
The electrical potential energy is given by;
P.E = 1/2 ke²
Where, k is the spring constant, e is the extension
Therefore;
Electrical potential energy = 1/2 × 113 N/m × 0.03 m
= 1.695 Joules
<u>= 1.7 J</u>
It might be radiation and reflection but I’m not sure
