Question

One method of identifying planets orbiting distant stars is by observing the planet passing in front of the star (known as a "transit"). You observe two of these transits exactly 247 days apart. Using the brightness and color of the star, you estimate its mass to be 8 × 1030 kg. At what distance from the star does the planet orbit? You will need to know the gravitational constant, G = 6.67x10-11 m3kg−1s−2. You may assume the planet orbits in a circle and is much lighter than the star.

Answer 1

r= 1.83 10¹¹ m

Step-by-step explanation:

For this exercise we use Newton's second law where the force is the force of universal attraction

F = m a

F = G m M / r²

acceleration is centripet

a = v² / r

we substitute

G m M /r² = m v² / r

G M /r = v2

Since the orbit is circular, the speed (modulus of speed) is constant, so we can use the uniform motion relation

v = d / t

the distance traveled in the orbit is equal to the length of the circle

d = 2π r

the time in this case is called period (T)

v = 2π r / T

we substitute

G M /r = 4π² r² / T²

r³ = G M T²/4π²

reduce the period to SI units

T = 247 d (24h / 1d) 3600 s / 1h) = 2,134 107 s

let's calculate

r³ = 6.67 10⁻¹¹ 8 10³⁰ (2,134 10⁷)²/ (4π²)

r =∛( 6.155 10³³)

r= 1.83 10¹¹ m

Answer 2

Given Information:

Time period = T = 247 days

Mass of planet = M = 8×10³⁰ kg

Required Information:

Distance from star = r = ?

Answer:

Distance from star = r = 1.833×10¹¹ m

Step-by-step explanation:

We know that time period is given by

`T = 2\pi \sqrt{(r^(3))/(GM) }`

Where r is the distance from the star that we want to find out, M is the mass of the planet and G is the gravitational constant.

G = 6.6743×10⁻¹¹ m³/kg⋅s²

`T = 2\pi \sqrt{(r^(3))/(GM) }\\\\(T)/(2\pi) = \sqrt{(r^(3))/(GM) }\\\\((T)/(2\pi))^(2) = (\sqrt{(r^(3))/(GM) })^(2) \\\\((T)/(2\pi))^(2) = (r^(3))/(GM) } \\\\r^(3) = ((T)/(2\pi))^(2) \cdot GM\\\\r = \sqrt[3]{(((T)/(2\pi))^(2) \cdot GM)}`

Convert time from days into seconds

Each day has 24 hours, each hour has 60 minutes, each minute has 60 seconds

T = 247*24*60*60

T = 2.134×10⁷ s

Substitute the given values

`r = \sqrt[3]{(((2.134* 10^(7))/(2\pi))^(2) \cdot 6.6743* 10^(-11)\cdot 8* 10^(30))}\\\\r = 1.833* 10^(-11) \: m`

Therefore, the planet is orbiting at a distance of 1.833×10¹¹ m from the star.