If the masses of both Jupiter and Saturn would suddenly increase 10 times, the force of gravity between them would:
3 answers
increase 100 times.
An equal-mass binary star system (M1 = M2) has an orbital period of 6 months and a semimajor axis of 2 au. What is the mass of each individual star of the binary?
To determine the mass of each individual star in the binary system, we can use Kepler's Third Law of Planetary Motion, which also applies to binary star systems.
Kepler's Third Law states that the square of the orbital period of a system is proportional to the cube of the semimajor axis:
T^2 = k * a^3
where T is the orbital period, a is the semimajor axis, and k is a constant.
In this case, we have an equal-mass binary star system, so we can assume that M1 = M2 = M (mass of each star). The total mass of the binary system can be expressed as:
Total Mass = 2 * M
Given that the orbital period is 6 months (or 0.5 years) and the semimajor axis is 2 au, we can plug these values into Kepler's Third Law to solve for the value of k:
(0.5)^2 = k * (2)^3
0.25 = 8k
k = 0.25/8 = 0.03125
Now, let's solve for the mass of each individual star. We know that the total mass equals 2 times the mass of each star:
Total Mass = 2 * M
Substituting the value of k and the orbital period into Kepler's Third Law, we get:
(0.5)^2 = 0.03125 * (2)^3 * M^2
0.25 = 0.03125 * 8 * M^2
0.25 = 0.25 * M^2
1 = M^2
Taking the square root of both sides, we find:
M = 1
Hence, the mass of each individual star in the binary system is 1 solar mass.
Kepler's Third Law states that the square of the orbital period of a system is proportional to the cube of the semimajor axis:
T^2 = k * a^3
where T is the orbital period, a is the semimajor axis, and k is a constant.
In this case, we have an equal-mass binary star system, so we can assume that M1 = M2 = M (mass of each star). The total mass of the binary system can be expressed as:
Total Mass = 2 * M
Given that the orbital period is 6 months (or 0.5 years) and the semimajor axis is 2 au, we can plug these values into Kepler's Third Law to solve for the value of k:
(0.5)^2 = k * (2)^3
0.25 = 8k
k = 0.25/8 = 0.03125
Now, let's solve for the mass of each individual star. We know that the total mass equals 2 times the mass of each star:
Total Mass = 2 * M
Substituting the value of k and the orbital period into Kepler's Third Law, we get:
(0.5)^2 = 0.03125 * (2)^3 * M^2
0.25 = 0.03125 * 8 * M^2
0.25 = 0.25 * M^2
1 = M^2
Taking the square root of both sides, we find:
M = 1
Hence, the mass of each individual star in the binary system is 1 solar mass.
1 answer