neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Physics

1 answer:

Vikentia [17]2 years ago

6 0

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

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Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv

lys-0071 [83]

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)