Question

A hypothetical planet has a mass of 1.66 times that of Earth, but the same radius. What is gravitiy near its surface?

Answer 1

Answer:

$g=16.28m/s^2$

Explanation:

The gravitational acceleration on the surface of the earth is

$g_{e}=\frac{Gm_{e}}{R_{e}^2}$

where G is the universal gravitational constant, $m_{e}$ is the mass of earth, and $R_{e}$ is the radius of earth,

in general for any object the gravitational acceleration or gravity on its surface is:

$g=\frac{Gm}{R^{2}}$

in this case we know that the mass is 1.66 times the mass of earth:

$m=1.66*m_{e}$

and the radius is the same as for earth:

$R=R_{e}$

so the gravity for this planet is

$g=\frac{G(1.66m_{e})}{R_{e}^2}$

which can be written in the following form:

$g=(1.66)\frac{Gm_{e}}{R_{e}^2}$

where we know that $g_{e}=\frac{Gm_{e}}{R_{e}^2}$ , so:

$g=(1.66)g_{e}$

and the acceleration of gravity on earth is: $g_{e}=9.81m/s^2$

so the acceleration or gravity on the planet is:

$g=(1.66)(9.81m/s^2)\\g=16.28m/s^2$

Answer 2

Answer: 16.22 m/s^2

Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so

((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2