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MrRissso [65]
2 years ago
10

R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d

etails of the equation; the basic point is that if you put an amount of mass M or larger in a sphere of radius smaller than RS(M)
What equation is this
Physics
1 answer:
padilas [110]2 years ago
3 0

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

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Explanation:

Given

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Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

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In probability:

P(X \ge 1) = 1 - P(X = 0)

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P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}

P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}

P(X=0) = (3)^0 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-3}

P(X=0) = e^{-3}

P(X=0) = 0.04979

Substitute 0.04979 for P(X=0) in P(X \ge 1) = 1 - P(X = 0)

P(X\ge 1) = 1 - 0.04979

P(X\ge 1) = 0.95021

P(X\ge 1) = 0.9502 ---  approximated

<em>Hence, the required probability is 0.9502</em>

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