Aplanet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass m 2.75 x 1029 kg. the radius of the orbit is r 8.05 x107 km. what is the orbital period (in earth days) of the planet tplanet?

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet

Mass of star

Radius of the orbit

Using centripetal and gravitational force

The centripetal force is given by

We know that,

....(I)

The gravitational force is given by

....(II)

From equation (I) and (II)

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

Hence, The orbital period of the planet is 387.62 days.

i think the answer is a, hope this : )

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s