Suppose our sun had 4 times its present mass but the earth orbited it at the same distance as it presently does. what would be the length of the year on the earth under those conditions? a.1/2 as long as the present yearb. twice as long as the present year. c. four times as long as the present year. d. the same as the present year. e. 1/4 as long as the present year

M₁ = initial mass of sun = M

M₂ = final mass of sun = 4 M

T₁ = initial time period

T₂ = final time period

r₁ = initial distance from sun = r

r₂ = final distance from sun = r

Using Kepler's third law

T₁² = 4π²r₁³/(GM₁)                                    eq-1

Using Kepler's third law

T₂² = 4π²r₂³/(GM₂)                                 eq-2

Dividing eq-1 by eq-2

T₁² /T₂² = (4π²r₁³/(GM₁))/(4π²r₂³/(GM₂))

T₁² /T₂² = (M₂/M₁) (r₁³/r₂³)

T₁² /T₂² = (4M/M) (r³/r³)

T₁² /T₂² = 4

T²₁ = 4 T²₂

Taking square root both side

T₁ = 2 T₂

T₂ = T₁ /2

hence the correct choice is

A. 1/2 as long as the present year

i was thinking more like reflects green light, just not sure

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2