A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from the Sun. If that star has sixteen times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period?

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

double the distance = divide strength by 4

b. a lightning bug, plug of grass, and a worm