Question
The principle of ‘parallax’ in section $2.3.1$ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit $\approx 3 \times 10^{11}m$. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of $1”$ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of $1”$ (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer

Diameter of Earth's orbit $=3 \times 10^{11} \mathrm{~m}$ Radius of Earth's orbit, $\mathrm{r}=1.5 \times 10^{11} \mathrm{~m}$ Let the distance parallax angle be1" $=$ $4.847 \times 10^{-6} \mathrm{rad}$. Let the distance of the star be D . Parsec is defined as the distance at which the average radius of the Earth's orbit subtends an angle of $1^{\prime \prime} \therefore$ We have $\theta=\frac{\mathrm{I}}{\mathrm{D}}$
$\text{D}=\frac{\text{r}}{\theta}=\frac{1.5\times10^{11}}{4.847\times10^{-6}}$
$=0.309\times10^{-6}\approx3.09\times10^{16}\text{m}$
Hence, 1 parsec $\approx3.09\times10^{16}\text{m}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Explain with the suitable example that a reversible process must be carried slowly and a fast process is necessarily irreversible.
A charged particle having a charge of $-2.0 \times 10^{-6} \mathrm{C}$ is placed close to a nonconducting plate having a surface charge density $4.0 \times 10^{-5} \mathrm{Cm}^{-2}$. Find the force of attraction between the particle and the plate.
The weight of a body at the poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about? Does your answer depend on whether only the earth's rotation is taken into account or the flattening of the earth at the poles is also taken into account?​​​​​
The speed of light in air is $3.00 \times 10^8 \mathrm{~ms}^{-1}$. The distance travelled by light in one year (i.e., 365 days $=3.154 \times 10^7 \mathrm{~s}$ ) is known as light year. A student calculates one light year $=9.462 \times 10^{15} \mathrm{~m}$. Do you agree with the student? If not, write the correct value of one light year.
Prove that the vectors $\vec{\text{A}}=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{B}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are mutuallly perpendicular.
Find the work done in pulling and pushing a roller through $100m$ horizontally when a force of $1500N$ is acting along a chain making an angle of 60° with ground. Assume the floor to be smooth.
What is the angular velocity at any point on the equator so that the body feels weightlessness?
Two bodies of m and 4m are placed at a distance. The gravitational field is zero at a point on the line joining the two masses. What will be the gravitational potential at this point?
Using the expression for pressure exerted by a gas, deduce Avogadro's law and Graham's law of diffusion.
Find an expression for the weight of a body at the centre of the Earth.