Respuesta :
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
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
Force of gravity on a spacecraft is 1748.5 N.
Explanation:
Given :
Distance = 12800 km
Mass = 1600 kg
Calculation :
Formula Used -
[tex]\rm F = \dfrac{GM_eM_s}{r^2}[/tex] ------ (1)
Where,
Radius of earth - [tex]\rm r = 1.91\times10^7\; m[/tex]
[tex]\rm M_s = 1600kg[/tex]
[tex]\rm M_e=5.98\times10^2^4\;kg\;\;\;\;(Mass \;of\;earth)[/tex]
[tex]\rm G = 6.67 \times 10^-^1^1\;Nm^2/kg^2[/tex]
So from equation (1),
[tex]\rm F= \dfrac{6.67\times10^-^1^1\times5.98\times10^2^4\times1600}{(1.91\times10^7)^2}[/tex]
[tex]\rm F = 1748.5\;N[/tex]
Therefore, force of gravity on a spacecraft is 1748.5 N.
For more information, refer the link given below
https://brainly.com/question/23271640?referrer=searchResults
