Question

A newly discovered planet is found to have density 2/3 Pe and radius 2Re, where Pe and Re are the density and radius of the Earth, respectively. The surface gravitational field of the planet is most nearly
A. 1.7 N/kg
B. 3.3 N/kg
C. 6.7 N/kg
D. 13 N/kg

Answer 1

Pls. see attachment for the answer online.

Answer 2

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'.

$g = G\frac{M_e}{R_e ^2}=9.81 N/kg\\ g'=G\frac{M_p}{R_p ^2}$

We know that the mass is given by product of volume and density. So,

$M_e = \rho_e \times \frac{4}{3} \pi R_e^3\\ M_p= \rho_p \times \frac{4}{3} \pi R_p^3$

It is given that,

$\rho_p = \frac{2}{3}\rho_e$

$R_p = 2 R_e$

$\Rightarrow M_p=\frac{2}{3}\rho_e \times \frac{4}{3} \pi (2R_e)^3= \frac{16}{3} M_e$

$g'=G\frac{16M_e}{3\times (2R_e)^2}=\frac{4}{3}\times G\frac{M_e}{R_e ^2}=\frac{4}{3}g = 13 N/kg$