Question: A space station includes a spherical observation dome with radius $ 3r $ and a conical observation tower with base radius $ r $ and height $ 4r $. What is the ratio of the volume of the cone to the sphere?

Sat, 02 May 2026 12:58:06 +0000

**Intro (Discover Hook)**
Imagine staring out from the far edge of a futuristic space station—wide-angle views of Earth glowing beneath the dome, framed by a sleek conical tower rising into the void. This iconic structure, where radius $ 3r $ defines the spherical observation dome and $ r $ and $ 4r $ shape a sharply tapered tower, raises more than just eyes—it sparks fascination. What’s the story behind these volumes? What’s the ratio of the cone’s space to the sphere’s, and why is understanding this ratio increasingly relevant in conversations about orbital architecture, design efficiency, and human presence in space? This precise question reflects growing interest in advanced space infrastructure and its physical foundations. **Why the Question Is Gaining Traction in the US**
Space architecture is no longer science fiction—it’s shaping transportation, tourism, and research industries worldwide. In the United States, where private aerospace firms and national space goals drive innovation, structures like this space station symbolize both progress and engineering precision. The visible contrast between the expansive spherical dome and the angular conical tower captures public interest, amplified by social media trends and digital storytelling around NASA collaborations, commercial space days, and orbital habitats. Today’s users seek clear, trustworthy insights into space design math—how shapes serve function, and why volume ratios matter in planning livable, efficient environments beyond Earth. **How the Volumes Are Calculated – A Neutral Explanation**
To find the ratio of the cone’s volume to the sphere’s, begin with the standard formulas. The sphere’s volume scales as $ V_s = \frac{4}{3}\pi R^3 $, with radius $ R = 3r $. So, $ V_s = \frac{4}{3}\pi (3r)^3 = \frac{4}{3}\pi (27r^3) = 36\pi r^3 $. The cone’s volume follows $ V_c = \frac{1}{3}\pi r^2 h $, with base radius $ r $ and height $ h = 4r $. Substituting, $ V_c = \frac{1}{3}\pi r^2 (4r) = \frac{4}{3}\pi r^3 $.