Question

The Earth orbits the Sun with a period of 3.16 x 107 seconds (1 year!) at a distance of 1.5 x 1011 m. The mass of the Earth is 6 x 1024 kg. Mars also orbits the Sun, with an orbital radius about 1.5 times larger than the radius of the Earths’s orbit. Based on this data, please find a) The mass of the Sun b) The length of a year on Mars

Answer 1

a) M = 2 10³⁰ kg , b)
`T_(m)` = 5.81 10⁷ s

Step-by-step explanation:

a) For this exercise let's use Newton's second law where force is the law of universal gravitation and acceleration is centripetal

G m M / R² = m a

a = v² / R

G M / R = V²

The orbit of the two planets is approximately circular, therefore the velocity module (speed) is constant

v = d / t

The distance is the length of the circular orbit

d = 2π R

G M / R = 4π² R² / T²

G M T² = 4π² R³

Let's write this equation for each planet

For the earth

The period is T = 3.16 10⁷ s and the radius of the orbit R = 1.5 10¹¹ m, let's calculate the mass of the sun

M = 4π² R³ / G T²

M = 4π² (1.5 10¹¹)³ / (6.67 10⁻¹¹ (3.16 10⁷)²)

M = 133.24 10³³ / 66.60 10³

M = 2 10³⁰ kg

b) For this part we write this equation for the two points

For the earth

`T_(E)`² = (4π² / G M)
`R_(E)`³

For mars

`T_(m)`² = (4π² / G M)
`R_(m)`³

Let's divide the two expressions

`T_(m)`² /
`T_(E)`² =
`R_(m)`³ /
`T_(E)`³

They indicate that the orbit of Mars is
`R_(m)` = 1.5
`R_(E)`

`T_(m)`² /
`T_(E)`² = (1.5
`R_(E)` /
`R_(E)`)³

`T_(m)`² =
`T_(E)`² 1.5³

`T_(m)`² = (3.16 10⁷)² 1.5³

`T_(m)` = √ (33.70 10¹⁴)

`T_(m)` = 5.81 10⁷ s