1. ## finding the radius problem

Hi i have a question: The radius of the Earth is approximately m and the acceleration due to gravity at the Earth’s surface is approximately 9.81 ms-2. If the acceleration due to gravity at the surface of the planet Mercury is approximately 3.59 ms-2 estimate the radius of Mercury. You may assume Newton’s Law of Gravitation but you do not require the value of the Gravitational constant in this law - list the other major assumptions that you make concerning the densities and shapes of the planets.

my attempt solution:
a = Gm/r²
for earth a = GM/R²
volume of sphere is ⁴/₃πr³
Assuming both have the same density
M = kR³
a = GM/R²
a = GkR³/R² = GkR

The ratio of the two values of a = 9.81/3.59 = 2.73
then the radius of mercury is 3.59=9.81*2.73R
gives R=0.13

is this incorrect? thx

2. Wait, for the last step, I'm not sure what you did exactly.

We have:

$a_{E} = GkR_{E}$

$a_{M} = GkR_{M}$

Then the ratio becomes:

$\frac{a_{E}}{a_{M}} = \frac{GkR_{E}}{GkR_{M}}$

This simplifies to:

$\frac{9.81}{3.59} = \frac{R_E}{R_M}$

You have to get R_M

$R_M = \frac{3.59R_E}{9.81}$

I don't know where your radius of the Earth went in your problem... I only see "radius of the Earth is approximately m"

EDIT: And why is latex not working again
EDIT2: seems okay now.

sorry for the mistype. The radius of the Earth is approximately 6.38*10^(-6) m

therefore to find Rm=[3.59*6.38]/9.81 = 2.34*10^(-6)m
is this correct? thanks!!

4. The figures are correct... but you got the wrong sign

10^-6 is in the realm of the micro world

5. sorry for the mistype AGAIN..lol The radius of the Earth is approximately 6.38*10^(6) m

therefore to find Rm=[3.59*6.38*(10^6)]/9.81 = 2335780.836m
thanks!!