Question

To what size would a star 5 times more massive than our Sun need

to be squished for it to become a black hole?




A.

15,000 m




B.

15 m




C.

15,000 kg




D.

15,000 km




E.

15 kg

Answer 1

Approximately
`15,\!000\; {\rm m}` in radius.

Step-by-step explanation:

For an object to become a black hole, the escape velocity from its surface should be at least as great as the speed of light in vacuum,
`c \approx 3.00* 10^(8)\; {\rm m\cdot s^(-1)}`.

The formula for escape velocity can be derived the fact that at escape velocity
`v`, the sum of the kinetic energy of an object
`\text{KE} = (1/2)\, m\, v^(2)` and its gravitational potential energy
`\text{GPE} = (-G\, M\, m) / (r)` is
`0`:

`\text{GPE} + \text{KE} = 0`.

`\displaystyle v = \sqrt{(2\, G\, M)/(r)}`,

Where:

• 
  `G \approx 6.67 * 10^(11)\; {\rm m^(3) \cdot kg^(-1)\cdot s^(-2)}` is the gravitational constant,
• 
  `M \approx 5 * (1.99 * 10^(30))\; {\rm kg}` is the mass of this star, and
• 
  `r` is the radius of this star.

Set escape velocity to the speed of light and solve for radius
`r`:

`\displaystyle c = \sqrt{(2\, G\, M)/(r)}`.

`\begin{aligned} r = (2\, G\, M)/(c^(2))\end{aligned}`.

Substitute in
`G \approx 6.67 * 10^(-11)\; {\rm m^(3)\cdot kg^(-1)\cdot s^(-2)}`,
`M \approx 5 * (1.99 * 10^(30))\; {\rm kg}`, and
`c \approx 3.00 * 10^(8)\; {\rm m\cdot s^(-1)}`:

`\begin{aligned} r &= (2\, G\, M)/(c^(2)) \\ &\approx (2\, (6.67 * 10^(-11))\, (5 * 1.99 * 10^(30)))/((3.00 * 10^(8))^(2)) \; {\rm m} \\&\approx 1.5 * 10^(4)\; {\rm m}\end{aligned}`.

In other words, the radius of this star should be no greater than
`1.5 * 10^(4)\; {\rm m}`.