Physics: The free-fall acceleration on the surface of planet a is about one tenth of that on the surface of planet b. if the radius of planet a is a quarter of the radius of planet b (rb/4), find the ratio of their average densities?

The free-fall acceleration on the surface of - id3869037, jasminecoronetti44

pamelperezz26

10.01.2022

$\frac{\rho_A}{\rho_B} = 0.4$

Explanation:

As we know that the acceleration due to gravity on the surface of the planet is given as

$a = \frac{GM}{R^2}$

so we will have

M = mass of the planet

$M = \rho (\frac{4}{3} \pi R^3)$

so we have

$a = \frac{G \rho \frac{4}{3} \pi R^3}{R^2}$

$a = \frac{4}{3}\rho \pi G R$

so we have

$\frac{a_A}{a_B} = \frac{\rho_A R_A}{\rho_B R_B}$

$\frac{1}{10} = \frac{\rho_A (R_B/4)}{\rho_B R_B}$

$\frac{1}{10} = \frac{\rho_A}{4\rho_B}$

$\frac{\rho_A}{\rho_B} = 0.4$

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emileep13

11.08.2022

you'd use the equation that establishes a relationship between energy and wavelength, which looks like this

e=h⋅cλ, where

λ - the wavelength of the photon; c - the speed of light, usually given as 3⋅108m/s; h - planck's constant, equal to 6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m

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