The planetMars requires 2.35 years to orbit
the sun, which has a mass of 2 × 1030 kg, in
an almost circular trajectory.
Find the radius of the orbit of Mars as it
circles the sun. The gravitational constant is
6.672 × 10−11 N · m2/kg2 .
Answer in units of m
Why did the sun and Mars go to couples therapy? Because they were having some serious space issues! But don't worry, I'll help you calculate the radius of Mars' orbit.
Let's start by using Kepler's Third Law which states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit. Mathematically, it can be represented as:
T^2 = (4π^2/GM) * a^3
Where T is the orbital period, G is the gravitational constant, M is the mass of the sun, and a is the radius of the orbit.
We're given that the orbital period of Mars is 2.35 years and the mass of the sun is 2 × 10^30 kg.
Substituting the values into the equation:
(2.35 years)^2 = (4π^2/6.672 × 10^−11 N·m^2/kg^2) * a^3
(2.35 * 365.25 days)^2 = (4 * π^2/6.672 × 10^−11 N·m^2/kg^2) * a^3
Doing the math, we get:
(2.35 * 365.25)² = (4 * π^2/6.672 × 10^−11) * a^3
Now, let's solve for a. Taking the cube root of both sides:
a = [(2.35 * 365.25)² * 6.672 × 10^−11)/(4 * π^2)]^(1/3)
Calculating that formula, the radius of Mars' orbit around the sun is approximately 2.28 * 10^11 meters (m).
So, Mars likes to keep its distance with a radius of approximately 2.28 * 10^11 meters away from the sun.
To find the radius of the orbit of Mars, we can use Kepler's Third Law of Planetary Motion, which relates the orbital period and the radius of the orbit.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit. In this case, since the orbit of Mars is almost circular, we can consider the radius (r) as the semi-major axis.
So, we have the equation:
T^2 = k * a^3,
where T is the orbital period, a is the radius, and k is a constant.
Given that the orbital period of Mars is 2.35 years (T = 2.35 years), we can substitute these values into the equation:
(2.35 years)^2 = k * a^3.
Now, we need to convert the orbital period from years to seconds, as the gravitational constant is given in SI units.
1 year = 365.25 days (taking into account the leap years) = 365.25 * 24 hours = 365.25 * 24 * 60 minutes = 365.25 * 24 * 60 * 60 seconds.
Substituting the value of the orbital period in seconds:
(2.35 years)^2 = k * a^3,
(2.35 * 365.25 * 24 * 60 * 60 seconds)^2 = k * a^3.
Now, we can solve for the constant k. The value of the gravitational constant (G) is 6.672 × 10^-11 N · m^2/kg^2, which can be written in the following form:
G = k * (mass of the sun) = k * (2 × 10^30 kg).
Rearranging the equation to solve for k:
k = G / (2 × 10^30 kg).
Now, we substitute the value of k into the equation:
(2.35 * 365.25 * 24 * 60 * 60 seconds)^2 = (G / (2 × 10^30 kg)) * a^3.
Simplifying the equation:
2.35^2 * (365.25 * 24 * 60 * 60 seconds)^2 = (G / (2 × 10^30 kg)) * a^3.
Finally, we solve for the radius (a) by taking the cube root of both sides:
a = [(2.35^2 * (365.25 * 24 * 60 * 60 seconds)^2) * (2 × 10^30 kg) / G]^(1/3).
Evaluating this expression will give us the radius of the orbit of Mars around the sun in meters.
forcegravity=centripetalforce
GMs*M/r^2=M w^2r
GMs/r^2=(2PI/period)^2 r
change period fromyears to seconds, look up G, solve for r