Question

The planet Mars has two moons. Phobos and Delmos (i) phobos has period 7 hours, 39 minutes and an orbital radius of 9.4×103km. Calculate the mass of Mars. (ii) Assume that Earth and mars move in a circular orbit around the sun, with the martian orbit being 1.52 times the orbital radius of the Earth. What is the length of the martian year in days? (G=6.67×10−11Nm2kg−2)

Answer 1

Explanation:

kepler law is used here

Answer 2

(I).  The mass of mars is $6.48\times10^{23}\ Kg$.

(II). The length of the martian yeas is 685 days.

Explanation:

Given that,

Period of Phobos = 7 hours 39 minutes

Orbital radius $r=9.4\times10^{3}\ km$

If m is the mass of Phobos and M is the mass of mars.

(I). We need to calculate the mass of mars

Using formula of centripetal force

$\dfrac{GmM}{r^2}=mr\omega^2$

$\dfrac{GmM}{r^2}=mr\times\dfrac{4\pi^2}{T^2}$

$M=\dfrac{4\pi^2\times r^3}{G\times T^2}$

Put the value into the formula

$M=\dfrac{4\pi^2\times(9.4\times10^{6})^3}{6.67\times10^{-11}\times(459\times60)^2}$

$M=6.48\times10^{23}\ Kg$

(II). If r₁,r₂ is the distance of Earth and Mars from the sun and T₁,T₂ are the periods of revolution of Earth and Mars around the sun, then

We need to calculate the length of the martian year in days

Using Kepler's third law

$T^2\propto r^3$

$T_{2}=(\dfrac{r_{2}}{r_{1}})^{\dfrac{3}{2}}\times T_{1}$

Put the value into the formula

$T_{2}=(1.52)^{\dfrac{3}{2}}\times365$

$T_{2}=684\ days$

Hence, (I).  The mass of mars is $6.48\times10^{23}\ Kg$.

(II). The length of the martian yeas is 685 days.

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Topic : Orbital radius

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