Geography: Titan has a radius of 2600.0 km and a mean density of 2.1 g/cm3. earth s moon has a radius of 1737.0 km and a mean density of 3.4 g/cm3. what is the ratio of gravitational acceleration on titan compared to that on the moon? the gravitational constant is g = 6.67 × 10-11 m3 kg-1 s-2.

5:1Step-by-step explanation:I m not positive on this but adding math and science would be 14 plus 11 plus 10, which is 35. 35 divided by 7 is 5...

Titan has a radius of 2600.0 km and a mean density

loganhowington26

06.01.2022

$\dfrac{g_1}{g_2}=0.92$

Explanation:

It is given that,

The radius of Titan, $r_1=2600\ km=26\times 10^5\ m$

Density of Titan, $d_1=2.1\ g/cm^3=2100\ kg/m^3$

Radius of moon, $r_2=1737\ km=1737\times 10^3\ m$

Density of the moon, $d_2=3.4\ g/cm^3=3400\ kg/m^3$

Value of gravitational constant, $G=6.67\times 10^{-11}\ m^3kg^{-1}s^{-2}$

The gravitational acceleration is given by :

$g=\dfrac{GM}{r^2}$

$g=\dfrac{G\times d\times \pi r^2}{r^2}$

Let g₁ and g₂ are the gravitational acceleration on Titan and on the moon respectively. So,

$\dfrac{g_1}{g_2}=\dfrac{\dfrac{G\times d_1\times (4/3)\pi r_1^3}{r_1^2}}{\dfrac{G\times d_2\times (4/3)\pi r_2^3}{r_2^2}}$

$\dfrac{g_1}{g_2}=\dfrac{d_1r_1}{d_2r_2}$

$\dfrac{g_1}{g_2}=\dfrac{2100\times 26\times 10^5}{3400\times 1737\times 10^3}$

$\dfrac{g_1}{g_2}=0.92$

So, the ratio of gravitational acceleration on Titan compared to that on the Moon is 0.92 : 1. Hence, this is the required solution.

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