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Venus has a diameter of 12,103.6 km (7,520.8 mi)—only 638.4 km (396.7 mi) less than Earth's—and its mass is 81.5% of Earth's, making it the third-smallest planet in the Solar System. Conditions on the Venusian surface differ radically from those on Earth because its dense atmosphere is 96.5% carbon dioxide, with most of the remaining 3.5% ...
*Venus [11 ] [12] [10] *Earth [13] ... at a larger size (900–1000 km diameter, rather than 400 km as for the moons of Saturn and Uranus). ... and is the distance ...
The surface of Venus is comparatively flat. When 93% of the topography was mapped by Pioneer Venus Orbiter, scientists found that the total distance from the lowest point to the highest point on the entire surface was about 13 kilometres (8.1 mi), about the same as the vertical distance between the Earth's ocean floor and the higher summits of the Himalayas.
The Venusian diameter of about 7,500 miles (12,000 km) is just a tad smaller than Earth's 7,900 miles (12,750 km). "Venus and Earth are often called sister planets because of their similarities in ...
This list contains a selection of objects 50 and 99 km in radius (100 km to 199 km in average diameter). The listed objects currently include most objects in the asteroid belt and moons of the giant planets in this size range, but many newly discovered objects in the outer Solar System are missing, such as those included in the following ...
In part, this is because Venus's dense atmosphere burns up smaller meteorites before they hit the surface. The Venera and Magellan data agree: there are very few impact craters with a diameter less than 30 kilometres (19 mi), and data from Magellan show an absence of any craters less than 2 kilometres (1 mi) in diameter. However, there are also ...
Venus currently has a surface temperature of 450℃ (the temperature of an oven’s self-cleaning cycle) and an atmosphere dominated by carbon dioxide (96%) with a density 90 times that of Earth’s.
This is related to the angular diameter distance, which is the distance an object is calculated to be at from and , assuming the Universe is Euclidean. The Mattig relation yields the angular-diameter distance, d A {\displaystyle d_{A}} , as a function of redshift z for a universe with Ω Λ = 0.