Arogue black hole with a mass 21 times the mass of the sun drifts into the solar system on a collision course with earth.

part a how far is the black hole from the center of the earth when objects on the earth's surface begin to lift into the air and "fall" up into the black hole?

give your answer as a multiple of the earth's radius. express your answer using three significant figures.

2640R where R = Radius of earth

Explanation:

Mass of Sun = 333000 of mass of Earth

Mass of rogue black hole (body) = 21 of mass of sun

So, mass of the body = 21 * 333000 of mass of Earth

Mass = 6993000 of mass of Earth

The gravitational attraction force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation distance

Mathematically,

M1 * (R2)² = M2 * (R1)²

Where M and R represents mass and radius

Make R2 the subject of the formula

R1 = SQRT ((M1 * R2²)/(M2))

R1 = R2 * SQRT(M1/M2)

Where R2 represents radius of Earth

M1 and M2 represents mass of object and mass of Earth respectively

M1 = 6993000* mass of Earth

So, equation becomes

R1 = R2 * SQRT(6993000 * M2/M2)

R1 = R2 * √6993000

R1 = 2644.4281R2

R1 = 2640R2 to 3 significant figures

So, the distance of the Black hole from the centre of the earth is 2640R

well clustered together.

points close on a scatter diagram