In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of 84 au. what is its orbital period according to kepler’s third law?

772.65 years.  

Explanation:

Semi major axis = R = 84 AU = 1.257 × 10¹³ m

According to the 3rd law of Kepler, T² =  4 π² R³ / GM

Here R is the semi major axis. G = 6.67 × 10⁻¹¹ SI units.

M is the mass of the  Sun  = 1.98 x 10³⁰ kg

T² =    4 (3.14)³ (1.257 × 10¹³ )³÷ (6.67 × 10⁻¹¹)(1.98 x 10³⁰)

  ⇒  Time period  =  T = 2.44 x 10¹⁰ seconds

1 s = 3.17 × 10⁻⁸ years

Converting the seconds to years, T = 772.65 years.

power out to power in less than 100%

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg