Respuesta :
Answers: (A)[tex]F=G\frac{M^2}{4R^2}[/tex] (B) [tex]V=\sqrt{\frac{GM}{4R}}[/tex] (C)[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex] (D)
[tex]E=-\frac{GM^{2}}{4R}[/tex]
Explanation:
(A) Gravitational force of one star on the other
According to the law of universal gravitation:
[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex] (1)
Where:
[tex]F[/tex] is the module of the gravitational force exerted between both bodies
[tex]G[/tex] is the universal gravitation constant.
[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.
[tex]r[/tex] is the distance between both bodies
In the case of this binary system with two stars with the same mass [tex]M[/tex] and separated each other by a distance [tex]2R[/tex], the gravitational force is:
[tex]F=G\frac{(M)(M)}{(2R)^2}[/tex] (2)
[tex]F=G\frac{M^2}{4R^2}[/tex] (3) This is the gravitational force between the two stars.
(B) Orbital speed of each star
Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed [tex]V[/tex] of each star is the same. In addition, if we assume this system is in equilibrium, gravitational force must be equal to the centripetal force [tex]F_{C}[/tex] (remembering we are talking about a circular orbit):
So: [tex]F=F_{C}[/tex] (4)
Where [tex]F_{C}=Ma_{C}[/tex] (5) Being [tex]a_{C}[/tex] the centripetal acceleration
On the other hand, we know there is a relation between [tex]a_{C}[/tex] and the velocity [tex]V[/tex]:
[tex]a_{C}=\frac{V^{2}}{R}[/tex] (6)
Substituting (6) in (5):
[tex]F_{C}=M\frac{V^{2}}{R}[/tex] (7)
Substituting (3) and (7) in (4):
[tex]G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}[/tex] (8)
Finding [tex]V[/tex]:
[tex]V=\sqrt{\frac{GM}{4R}}[/tex] (9) This is the orbital speed of each star
(C) Period of the orbit of each star
The period [tex]T[/tex] of each star is given by:
[tex]T=\frac{2\pi R}{V}[/tex] (10)
Substituting (9) in (10):
[tex]T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}[/tex] (11)
Solving and simplifying:
[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex] (12) This is the orbital period of each star.
(D) Energy required to separate the two stars to infinity
The gravitational potential energy [tex]U_{g}[/tex] is given by:
[tex]U_{g}=-\frac{Gm_{1}m_{2}}{r}[/tex] (13)
Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy [tex]\Delta U_{g}[/tex] for this case is:
[tex]\Delta U_{g}=U-U_{\infty}[/tex] (14)
Knowing [tex]U_{\infty}=0[/tex] the total potential energy is [tex]U[/tex] and in the case of this binary system is:
[tex]U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}[/tex] (15)
Now, we already have the potential energy, but we need to know the kinetic energy [tex]K[/tex] in order to obtain the total Mechanical Energy [tex]E[/tex] required to separate the two stars to infinity.
In this sense:
[tex]E=U+K[/tex] (16)
Where the kinetic energy of both stars is:
[tex]K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}[/tex] (17)
Substituting the value of [tex]V[/tex] found in (9):
[tex]K=M(\sqrt{\frac{GM}{4R}})^{2}[/tex] (17)
[tex]K=\frac{1}{4}\frac{GM^{2}}{R}[/tex] (18)
Substituting (15) and (18) in (16):
[tex]E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}[/tex] (19)
[tex]E=-\frac{GM^{2}}{4R}[/tex] (20) This is the energy required to separate the two stars to infinity.
F = [tex]\frac{GM1M2}{R} \\\\[/tex]
Where F = Gravitational force that is between the two masses
M1 = mass of the first star
M2 = Mass of the second star
d= distance between the masses
G = Gravitational constant
These stars are identical so M1 = M2 = M
d = 2R
d² = 2R² = 4R²
[tex]F= \frac{GM1M2}{4R^2}[/tex]
B. The orbital speed
Given that theses circles are said to move in a circular orbit, the net centripetal force
Fnet = F second law of Newton
[tex]Fnet = \frac{MV^2}{R} =\frac{GM^2}{4R^2} \\\\v^2 = \frac{GM}{4R}[/tex]
[tex]v = \sqrt[]{}\frac{{GM} }{4R}[/tex]
C. The orbital period
Period = time T
speed = distance traveled / time
distance = circumference of a circle = 2πR
V = 2πR/T
T = 2πR/V
[tex]T = 2\pi R(\sqrt{(MG)/(4R)}^{-1} \\[/tex]
[tex]T = (2\pi R\sqrt{4R} )/(\sqrt{MG} )[/tex]
[tex]T = (2\pi R(2R^{1/2} ))/\sqrt{MG}[/tex]
[tex]T = (4\pi R^{3/2}) /\sqrt{MG[/tex]
Read more on orbital speed on https://brainly.com/question/22247460
