Equal solid angle by two unequal spheres

Geometry Level 5

Two unequal spheres having radii R 1 R_1 & R 2 R_2 are at distances D 1 = 7 km D_1=7 \text{km} & D 2 = 100 km D_2=100 \text{km} respectively measured from their centers to a given point in the space. It is found that both the spheres ( R 1 ≠ R 2 R_1\ne R_2 ) subtend equal solid angle at that point. Find out the ratio R 1 / R 2 R_1/ R_2 .


The answer is 0.07.

Harish Chandra Rajpoot by Harish Chandra Rajpoot , 30, India

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1 solution

Solid angle subtended by a sphere of radius r r at any point in the space lying at a distance d d from the center of sphere is given as Ω = 2 π ( 1 − cos ⁡ α ) = 2 π ( 1 − d 2 − r 2 d ) \Omega=2\pi(1-\cos\alpha)=2\pi \left(1-\frac{\sqrt{d^2-r^2}}{d}\right)

since both the unequal spheres subtend equal solid angle at the given point in the space hence we have

2 π ( 1 − D 1 2 − R 1 2 D 1 ) = 2 π ( 1 − D 2 2 − R 2 2 D 2 ) ⟹ R 1 R 2 = D 1 D 2 2\pi \left(1-\frac{\sqrt{D_1^2-R_1^2}}{D_1}\right)=2\pi \left(1-\frac{\sqrt{D_2^2-R_2^2}}{D_2}\right)\implies \frac{R_1}{R_2}=\frac{D_1}{D_2}

plugging the values of D 1 D_1 & D 2 D_2 , R 1 R 2 = 7 100 = 0.07 \frac{R_1}{R_2}=\frac{7}{100}=0.07

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Ω = 2 π ( 1 − cos ⁡ α ) = 2 π ( 1 − d 2 − r 2 d ) \Omega=2\pi(1-\cos\alpha)=2\pi \left(1-\frac{\sqrt{d^2-r^2}}{d}\right)

Where did you get this formula?

Pi Han Goh - 4 years, 5 months ago

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Nice Question, consider an imaginary conical surface (i.e. conical shell) of minimum apex angle 2 α 2\alpha having its vertex at the given point in the space such that it exactly encloses a sphere of radius r r at a central distance d d from the given point then the solid angle Ω \Omega subtended by the sphere at the given point

Ω = solid angle subtended by the conical surface at its vertex = 2 π ( 1 − cos ⁡ α ) \Omega=\text{solid angle subtended by the conical surface at its vertex}=2\pi(1-\cos\alpha)

Join the given point to the center of sphere & draw a tangent to the sphere to get the value of cos ⁡ α \cos \alpha as follows

cos ⁡ α = d 2 − r 2 d \cos \alpha=\frac{\sqrt{d^2-r^2}}{d} Thus you get the final expression of solid angle by a sphere at any point in the space

Harish Chandra Rajpoot - 4 years, 5 months ago

@Pi Han Goh : Nice Question, consider an imaginary conical surface (i.e. conical shell) of minimum apex angle 2 α 2\alpha having its vertex at the given point in the space such that it exactly encloses a sphere of radius r r at a central distance d d from the given point then the solid angle Ω \Omega subtended by the sphere at the given point

Ω = solid angle subtended by the conical surface at its vertex = 2 π ( 1 − cos ⁡ α ) \Omega=\text{solid angle subtended by the conical surface at its vertex}=2\pi(1-\cos\alpha)

Join the given point to the center of sphere & draw a tangent to the sphere to get the value of cos ⁡ α \cos \alpha as follows

cos ⁡ α = d 2 − r 2 d \cos \alpha=\frac{\sqrt{d^2-r^2}}{d}

Harish Chandra Rajpoot - 4 years, 5 months ago

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