Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth). What must be the mass of this planet such that the gravitational force (Fgravity) at the surface would be identical to Earth's?

Question

contestada

Respuesta :

boffeemadrid boffeemadrid · Answer 1 · 2019-12-11T07:03:48+01:00

Answer:

4 times the mass of Earth

Explanation:

[tex]M_1[/tex] = Mass of Earth

[tex]M_2[/tex] = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

[tex]F=\frac{GM_1m}{r^2}[/tex]

The force of gravity on an object on the other planet is

[tex]F=\frac{GM_2m}{(2r)^2}[/tex]

As the forces are equal

[tex]\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1[/tex]

So, the other planet would have 4 times the mass of Earth

andromache andromache · Answer 2 · 2022-02-15T15:07:06+01:00

The mass of this planet should be 4 times the mass of Earth.

Here

M_1 means the mass of the earth

M_2 Mass of the other planet

r be the Radius of Earth

2r be the Radius of the other planet

And, m be the Mass of object

Now

We know that

[tex]\frac{GM_1m}{r^2}= \frac{GM_2m}{(2r)^2} \\\\M_1 = \frac{M_2}{4}\\\\ M_2 = 4M_1[/tex]

Learn more about the mass here: https://brainly.com/question/14511491