Respuesta :
Answer: D. 13 N/kg
Explanation:
The surface gravitational field is given by acceleration due to gravity on the planet (g). Let the value of it on a newly discovered planet be g'.
[tex]g = G\frac{M_e}{R_e ^2}=9.81 N/kg\\ g'=G\frac{M_p}{R_p ^2}[/tex]
We know that the mass is given by product of volume and density. So,
[tex]M_e = \rho_e \times \frac{4}{3} \pi R_e^3\\ M_p= \rho_p \times \frac{4}{3} \pi R_p^3[/tex]
It is given that,
[tex]\rho_p = \frac{2}{3}\rho_e[/tex]
[tex]R_p = 2 R_e[/tex]
[tex]\Rightarrow M_p=\frac{2}{3}\rho_e \times \frac{4}{3} \pi (2R_e)^3= \frac{16}{3} M_e[/tex]
[tex]g'=G\frac{16M_e}{3\times (2R_e)^2}=\frac{4}{3}\times G\frac{M_e}{R_e ^2}=\frac{4}{3}g = 13 N/kg[/tex]
The surface gravitational field of the planet is most nearly; D: 13 N/kg
Gravitational Field
We are given;
- Density of new planet; ρ = ²/₃ρ_e
- Radius of new planet; R = 2R_e
Formula for density is;
ρ = m/V
where;
m is mass
V is volume
Thus, for the new planet, we can calculate the mass of the planet from;
M_p = ρ × V
Planet is spherical in shape and as such the volume is;
V = ⁴/₃πr³
Thus;
M_p/V_p = ²/₃(M_e/V_e)
M_p/(⁴/₃πR³) = ²/₃(M_e/(⁴/₃πR_e³))
⁴/₃π will cancel out to give;
M_p/R³ = ²/₃(M_e/R_e³)
M_p = ²/₃R³(M_e/R_e³)
Formula for gravitational field of the new planet is;
g' = GM/r²
Thus;
g' = G[²/₃R³(M_e/R_e³)]/(2R_e)²
⇒ g' = ⁴/₃GM_e/R_e²
Now, GM_e/R_e² is equal to acceleration due to gravity = 9.8 m/s². Thus;
g' = ⁴/₃ × 9.8
g' ≈ 13 m/s²
But the units are in N/kg and converting to N/kg gives us;
g' = 13 N/kg
Read more about gravitational field at; https://brainly.com/question/14080810
