56.5k views
0 votes
Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from mars to the sun, then calculate it

User SBirthare
by
8.0k points

1 Answer

1 vote
This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is

v=\frac{2\pi{r}}{T}

=\frac{2\pi{r}}{1.88*365.25*86400} m/s, accounting for leap years

=3.371*10^(-8)\pi{r} m/s

The centripetal force, Fc, generated is

Fc=(mv^2)/(r)
where m=mass of mars = 6.39*10^(24) kg

=(mv^2)/(r)

=(6.39*10^(24)v^2)/(r)

=7.26168*10^9\pi^2r

The gravitation pull from the sun, Fg, is given by

Fg=(GMm)/(r^2)
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg

=(6.67408*10^(-11)1.989*10^(30)6.39*10^(24))/(r^2)

=(8.4826*10^44)/(r^2)

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>

7.26168*10^9\pi^2r=(8.4826*10^44)/(r^2)
Solving for the real root:

r^3=(8.48256*10^44)/(7.26168*10^9*%pi^2)

=(1.1681263*10^(35))/(\pi^2)
=2.279*10^11 m


User Deepak Goyal
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.