Question

A planet of mass m = 1.55 x 10²⁴ kg is orbiting in a circular path a star of mass M = 9.75 x 10²⁹ kg. The radius of the orbit is R = 4.65 x 10⁷ km. What is the orbital period (in earth days) of the planet? Where G = 6.67 � 10^-11 N�m²/kg² and 1 day = 8.54 � 10⁴ s.

I used the formula
$\frac{T^2}{r^3} = \frac{4\pi^2}{GM}$ and got T = 247 s = 0.0289 days, but the online homework system says thats the wrong answer.

What am I doing wrong?

Answer 1

T = 2*pi sqrt(r^3 / GM)
R=4.65*10^7km= 4.65*10^10metre

M=mass of star=9.75 x 10²⁹ kg

T=7686902.828 sec or 90.01 days make sure u converted the radius in metres and took mass of star

Answer 2

The gravitational potential energy is equal to the centripetal force GmM mi 7 7" GM GM 4T273 GM GM Substitute the given values. (4.65x10' km1 6.67 x10-11 N m/kg2)(9.75x1029 kg 1 day (7.80%10860012) 7.80x106 s 90 days Therefore, the orbital period is 90 days

Answer 3

T = 2*pi sqrt(r^3 / GM)
R=4.65*10^7km= 4.65*10^10metre

M=mass of star=9.75 x 10²⁹ kg

G= 6.67