What is the percentage or ratio of strength of gravity between earth and a blackhole?

Question

For example, earth's gravity is two trillion times weaker than a black hole then put that in a percentage.ie. 000000000002

Put it another way. The amount of gravity needed to prevent light from escaping.

compare that to earths.

I think the answer will be similar to the excape velocity.

So the speed of graivty at a blackhole is 27,272 times quicker than earths?

11(earth's escape vel)x27,272=300000km/s aprox

Answer 1

OK, your actual question is probably "what is the ratio between the escape velocity on the surface of earth and the escape velocity on the surface of a blackhole", where gravity from all other sources such as the Sun are ignored. Your answer seems approximately correct. Let's double check, letting the answer be x.
x = Ve_Earth_Surface / Ve_Blackhole_Surface

Ve_Earth_Surface = 11.2 km/s. So,
x = 11.2 / Ve_Blackhole_Surface

Ve_Blackhole_Surface = speed of light = c =~ 299,792 km/s. So,
x = 11.2 / 299,792
=~ 0.00374%, or a ratio of 1:26767

Note1) Proof for "Ve_Blackhole_Surface = c":
Escape velocity = Ve = (2*G*M/r)^0.5
where G = gravitational constant
M = mass of the black hole
r = radius on the surface
= the "Schwarzschild radius" for a black hole = 2*G*M/(c^2) where c = speed of light
So,
Ve_Blackhole_Surface = (2*G*M/ {2*G*M/c^2})^0.5 = (c^2)^0.5 = c

Note 2) If x was Ve_Earth_Surface / Ve_Inside_Blackhole then x = 0, since Ve_Inside_Blackhole is infinity because nothing can escape directly from inside a blackhole (Hawking Radiation notwithstanding).

Note 3) For distances further away from the blackhole surface, Ve_Blackhole depends on the blackhole's mass and the distance, so the answer becomes:
f(r) = 11.2 / Ve_Blackhole(r), where r = distance away from the center of the blackhole, assuming r > the Schwarzschild radius.
Then,
f(r) = 11.2 / (2*G*M/r)^0.5

Answer 2

You could put a figure to this question but you would not take into account that once it is a blackhole the gravitational pull is preventing light to escape so we know it has reached a mass capable of this.
But what happens now if the mass increases?
The gravitational pull would as a result increase.
Therefore any calculations derived from the comparitive escape velocity would determine the minimum gravitational pull of a blackhole. Since also they are increasing in mass they would also be increasing in gravitational pull. And considering the distance of suspected blackholes we can only assume that what we observe is old because of how long it took light to get here (also consider that the blackhole would have affected this because of its ability to warp light ). So even for each individual blackhole (each of differing mass) we'd only assume they'd be as unique as the stars they formed from.
In short perhaps it is currently an impossibility to calculate the actual power of gravity on any single blackhole. Whatever figure we determine as the minimum gravitational pull is the all we can actually determine.

But as a comparison noted scientists have discussed these matters and based on their knowledge I have heard discussion that to create the same gravitational effect of earth the actual size of a blachhole would need be no larger than a golf ball. The ratio of size is enormous as you know.

Answer 3

Too vague. Gravitational pull where? At the surface? At what distance? And do you have a specified mass for this black hole? You can make a black hole out of pretty much anything if you compact it down enough, but making a black hole out of Pluto wouldn't have that much of a gravitational effect on anything - not more than it does now, anyway. If the Sun became a black hole tomorrow, the Earth's orbit wouldn't change.

Answer 4

The strength of an object's gravitational field varies with the distance away from that object - i.e., as you get farther away from it, the gravitational field strength weakens. It also varies with the mass of the object. So to properly answer your question, we need to know the mass of the black hole (they come in every mass imaginable) and the distance away from the black hole and the distance away from Earth you are interested in.

For instance, say we have a black hole with the same mass as Earth. The radius of Earth is about 4000 miles. So the gravitational field strength at the surface of Earth is the same as the strength 4000 miles from the center of the black hole.

Answer 5

interior the midsection of the Milky way is a extensive decision of stars. interior the midsection is assumed to be a extensive black hollow. the subsequent closest danger is Canis important, approximately 25,000 mild-years away.