Respuesta :
Answer:
a) M = 20,327,895.776 solar masses = 20.33 million solar masses
b) No, this massive object cannot be a single, ordinary star
c1) The Schwarzschild radius of this black hole will be R = 59.96 * 10⁹ m
c2) No, a black hole of this size will not fit inside the earth's orbit around the sun
Explanation:
Let the Mass of the object at the cengter of the milky way galaxy be = M
Centripetal Force, [tex]F_{c} =\frac{mv^{2} }{r}[/tex]
Gravitational force, [tex]F_{g} = G\frac{Mm}{r^{2} }[/tex]
Equating [tex]F_{c} and F_{g}[/tex]
[tex]\frac{mv^{2} }{r} = G\frac{Mm}{r^{2} } \\M = \frac{v^{2}r }{G}[/tex]
v = 190 km/s = 190000 m/s
Diameter, d = 15 light years
Radius, r = 15/2 = 7.5 light years
1 light year = 9.461 * 10¹⁵m
r = 7.5 * 9.461 * 10¹⁵ m = 74.7419 * 10¹⁵ m
G = 6.67 * 10⁻¹¹
[tex]M = \frac{190000^{2} *74.7419 * 10^{15} }{6.67 * 10^{-11} }[/tex]
M = 40.45 * 10³⁶ kg
1 solar mass = 1.99 * 10³⁰ kg
M [tex]= \frac{40.45 * 10^{36} }{1.99 * 10^{30} }[/tex]
M = 20,327,895.776 solar masses
Mass of the object is about 20.33 million solar masses
b) This object cannot be a single, ordinary star since the mass which is approximately 20.33 million solar masses is more than the mass possible for a single star. i.e less than 50 solar masses
c1) The radius of the black hole can be given by the relation
[tex]C = \sqrt{\frac{2GM}{R} } \\R = \frac{2GM}{C^{2} }[/tex]
The speed of light, C = 3 * 10⁸ m/s
[tex]R = \frac{2 * 6.67 * 10^{-11}*40.45 * 10^{36} }{( 3*10^{8}) ^{2} }[/tex]
R = 59955888888.89 m
R = 59.96 * 10⁹ m
c2) No, a black hole of this size will not fit inside the earth's orbit around the sun because the radii of the earth and the sun are 6,371 km and 696,340 km respectively, which are far less than the radius of this black hole
Answer:
A) In Kg, M = 3.84 x 10^(37) kg ; In solar masses, M = 1.931 x 10^(7) solar masses
B) No, it can't be a single ordinary star.
C)Schwarzschild Radius = 5.69 x 10^(10)m
- Yes, a black hole of the object size can fit into the earths orbital radius.
Explanation:
A) Let's assume that the ring shaped disk is circular and thus, to find the mass of the object at the centre of the milky way, we can use the circular orbit equation;
v = √(GM/r)
where;
G is the gravitational constant and has a value of 6.67 x 10^(-11) Nm²/kg²
R is the radius of the orbit,
M is the mass of the larger object
v = velocity = 190km/s = 190000 m/s
The question says it has a diameter of 15 light years. Thus, radius = 15/2 = 7.5 light years
Now, let's convert it to metres.
1 light year = 9.4605 x 10^(15) m
Thus,7.5 light years = 7.5 x 9.4605 x 10^(15) m
Making M the subject of the formula;
M = v²r/G
Plugging all the relevant values, we have;
M = [190000² x 7.5 x 9.4605 x 10^(15)]/6.67 x 10^(-11)
M = 3.84 x 10^(37) kg
The question says we should also express this value in solar mass.
To do this, we'll just divide this mass by the mass of the sun.
Now, mass of the sun generally has a value of 1.9891 x 10^(30) kg
Thus, value of mass of object in solar masses = 3.84 x 10^(37)/1.9891 x 10^(30) = 1.931 x 10^(7) solar masses
B) No, it cant be a single ordinary star because this object has a mass of 1.931 x 10^(7) solar masses and since it is impossible for a single star to have a mass of more than about 50 solar masses, the mass of the object is far greater than the minimum mass of a start and thus the object cannot be a single ordinary star.
C) The formula for Schwarzschild radius is given as;
R_s = 2GM/c²
Where c is speed of light = 3 x 10^(8) m/s
Thus,
R_s = 2 x 6.67 x 10^(-11) x 3.84 x 10^(37)/(3 x 10^(8))² = 5.69 x 10^(10)m
To find out if a black hole of this size fit inside the earth’s orbit around the sun.
Radius of Earths orbit (R_e) is given as;
R_e = 149.6 x 10^(6) km = 14.9 x 10^(10) m
Comparing R_s and R_e, we see that R_s is lesser than R_e and thus, the object can fit into the earths orbital radius.
