Question

Alpha-Centari, is located at a distance of 4.5 light years. What is the diameter of a reflecting telescope, placed in Earths orbit, observed in visible light at 550nm, in order to resolve:a-A planet orbiting its star at the same distance as earth is from the sun.b- A sun sized star. c- Earth sized planet?

Answer 1

Alpha-Centari, is located at a distance of 4.5 light years. What is the diameter of a reflecting telescope, placed in Earths orbit, observed in visible light at 550nm, in order to resolve:

a-A planet orbiting its star at the same distance as earth is from the sun.

b- A sun sized star.

c- Earth sized planet?

Solution

According the theory of difraction the angular resolution of telescope with a objective diameter $D$ is

$$\alpha = 1.22 * \frac{\lambda}{D} * 206265' \text{ (in angular seconds)}$$

Where $\lambda$ is wavelength of light

a) We have the angular distance between star and planet, if a planet orbiting its star at the same distance as earth is from the sun and star is located at a distance of 4.5 light years= 284582 astronomical unit, is

$$\alpha = \frac{1\text{a.u.}}{284582\text{a.u.}} * 206265' \approx 0.725'$$

From hence we have minimal diameter od telescope

$$D_{(a)\min} = \frac{1.22 * \lambda * 206265'}{\alpha} = \frac{1.22 * 550\text{nm} * 206265'}{\alpha} = 191\text{mm} \Rightarrow D_a > 191\text{mm}$$

b) We have the angular size of star, if a star is like Sun (star radius is $6,9551 \cdot 10^8$ m) and star is located at a distance of 4.5 light years= 284582 astronomical unit $\approx 4.26 \cdot 10^{16}$ m, is

$$\alpha = \frac{6,9551 \cdot 10^8 \text{m}}{4.26 \cdot 10^{16} \text{m}} * 206265' \approx 0.0033'$$

From hence we have minimal diameter od telescope

$$D_{(b)\min} = \frac{1.22 * 550\text{nm} * 206265'}{0.0033'} \approx 41962\text{mm} \approx 42\text{m} \Rightarrow D_b > 42\text{m}$$

c) We have the angular size of planet, if a planet is like Earth (Earth radius is $6,4 \cdot 10^6$ m) and star is located at a distance of 4.5 light years= 284582 astronomical unit $\approx 4.26 \cdot 10^{16}$ m, is

$$\alpha = \frac {6 . 4 1 0 ^ {6} \mathrm {m}}{4 . 2 6 \cdot 1 0 ^ {1 6} \mathrm {m}} * 2 0 6 2 6 5 ^ {\prime} \approx 0. 0 0 0 0 3 ^ {\prime}$$

From hence we have minimal diameter od telescope

$$D _ {(c) \min } = \frac {1 . 2 2 * 5 5 0 n m * 2 0 6 2 6 5 ^ {\prime}}{0 . 0 0 0 0 3 ^ {\prime}} \approx 4 2 0 0 m \Rightarrow$$ $$D _ {c} > 4 2 0 0 m$$

**Answer**

a)

$$D _ {(a) \min } = \frac {1 . 2 2 * \lambda * 2 0 6 2 6 5 ^ {\prime}}{\alpha} = \frac {1 . 2 2 * 5 5 0 n m * 2 0 6 2 6 5 ^ {\prime}}{\alpha} = 1 9 1 m m \Rightarrow$$ $$D _ {\alpha} > 1 9 1 m m$$

b)

$$D _ {(b) \min } = \frac {1 . 2 2 * 5 5 0 n m * 2 0 6 2 6 5 ^ {\prime}}{0 . 0 0 3 3 ^ {\prime}} \approx 4 1 9 6 2 m m \approx 4 2 m \Rightarrow$$ $$D _ {b} > 4 2 m$$

c)

$$D _ {(c) \min } = \frac {1 . 2 2 * 5 5 0 n m * 2 0 6 2 6 5 ^ {\prime}}{0 . 0 0 0 0 3 ^ {\prime}} \approx 4 2 0 0 m \Rightarrow$$ $$D _ {c} > 4 2 0 0 m$$