Respuesta :
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
Explanation:
We know that time period is given by
[tex]T = 2\pi \sqrt{\frac{r^{3}}{GM} }[/tex]
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
[tex]T = 2\pi \sqrt{\frac{r^{3}}{GM} }\\\\\frac{T}{2\pi} = \sqrt{\frac{r^{3}}{GM} }\\\\(\frac{T}{2\pi})^{2} = (\sqrt{\frac{r^{3}}{GM} })^{2} \\\\(\frac{T}{2\pi})^{2} = \frac{r^{3}}{GM} } \\\\r^{3} = (\frac{T}{2\pi})^{2} \cdot GM\\\\r = \sqrt[3]{((\frac{T}{2\pi})^{2} \cdot GM)}[/tex]
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
[tex]r = \sqrt[3]{((\frac{2.134\times 10^{7}}{2\pi})^{2} \cdot 6.6743\times 10^{-11}\cdot 8\times 10^{30})}\\\\r = 1.833\times 10^{-11} \: m[/tex]
Therefore, the planet is orbiting at a distance of 1.833×10¹¹ m from the star.
Answer:
r= 1.83 10¹¹ m
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
