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Using Radiation to Investigate Stars

The inverse square law, stellar distance and multiwavelength astronomy

  • 3 things to know
  • Component 2

Overview

Recall questions on how intensity falls with distance, using the luminosity and the measured intensity to find a distance, worked calculations, and why observations in more than one region of the electromagnetic spectrum reveal what visible light alone cannot.

What you need to know

Answer each one before you reveal it.

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Key term

State the inverse square law for the intensity of radiation received from a star, defining every symbol and its unit.

Key term

Why does the radiation received from a star obey an inverse square law?

Key term

By what factor does the intensity received fall if the distance to a star is trebled?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the inverse square law for the intensity of radiation received from a star, defining every symbol and its unit.

    Model answer

    The intensity is I = L/(4πd²), where I is the intensity in W m⁻², L is the star's luminosity in watts and d is its distance from the observer in metres.

  2. 2

    Why does the radiation received from a star obey an inverse square law?

    Model answer

    The star radiates its power equally in all directions, so at a distance d that power is spread over the surface of a sphere of area 4πd², and the power per square metre falls as the inverse square of the distance.

  3. 3

    What assumption about the space between a star and the observer does the inverse square law make?

    Model answer

    It assumes the intervening space is empty, so none of the radiation is absorbed or scattered before it arrives.

  4. 4

    By what factor does the intensity received fall if the distance to a star is trebled?

    Model answer

    The intensity falls to one ninth of its value, because it is inversely proportional to the square of the distance.

  5. 5

    Calculate the intensity of solar radiation at the Earth, taking the Sun's luminosity as 3.9 × 10²⁶ W and the Earth's distance as 1.50 × 10¹¹ m.

    Model answer

    The intensity is I = 3.9 × 10²⁶ ÷ (4π × (1.50 × 10¹¹)²) = 1.4 × 10³ W m⁻². Multiplying 1380 W m⁻² by 4π × 2.25 × 10²² m² returns 3.9 × 10²⁶ W.

  6. 6

    A star of luminosity 4.0 × 10²⁸ W lies 3.0 × 10¹⁸ m away. Calculate the intensity of its radiation at the Earth.

    Model answer

    The intensity is I = 4.0 × 10²⁸ ÷ (4π × (3.0 × 10¹⁸)²) = 3.5 × 10⁻¹⁰ W m⁻². Multiplying 3.54 × 10⁻¹⁰ W m⁻² by 1.131 × 10³⁸ m² returns 4.0 × 10²⁸ W.

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