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A common misconception occurs between centre of mass and centre of gravity.They are defined in similar ways but are not exactly the same quantity. Centre of mass is the mathematical description of placing all the mass in the region considered to one position, centre of gravity is a real physical quantity, the point of a body where the gravitational force acts.
According to the IAU's explicit count, there are eight planets in the Solar System; four terrestrial planets (Mercury, Venus, Earth, and Mars) and four giant planets, which can be divided further into two gas giants (Jupiter and Saturn) and two ice giants (Uranus and Neptune). When excluding the Sun, the four giant planets account for more than ...
The percentage columns show the distance from the orbit compared to the semimajor axis. E.g. for the Moon, L 1 is 326 400 km from Earth's center, which is 84.9% of the Earth–Moon distance or 15.1% "in front of" (Earthwards from) the Moon; L 2 is located 448 900 km from Earth's center, which is 116.8% of the Earth–Moon distance or 16.8% ...
The standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of that body. For two bodies, the parameter may be expressed as G ( m 1 + m 2 ) , or as GM when one body is much larger than the other: μ = G ( M + m ) ≈ G M . {\displaystyle \mu =G(M+m)\approx GM.}
Entering a Hohmann transfer orbit from Earth to Jupiter from low Earth orbit requires a delta-v of 6.3 km/s, [170] which is comparable to the 9.7 km/s delta-v needed to reach low Earth orbit. [171] Gravity assists through planetary flybys can be used to reduce the energy required to reach Jupiter. [172]
The actual Hill radius for the Earth-Moon pair is on the order of 60,000 km (i.e., extending less than one-sixth the distance of the 378,000 km between the Moon and the Earth). [9] In the Earth-Sun example, the Earth (5.97 × 10 24 kg) orbits the Sun (1.99 × 10 30 kg) at a distance of 149.6 million km, or one astronomical unit (AU). The Hill ...
For example, the fact that Earth is a gravitationally-bound sphere of its current size costs 2.494 21 × 10 15 kg of mass (roughly one fourth the mass of Phobos – see above for the same value in Joules), and if its atoms were sparse over an arbitrarily large volume the Earth would weigh its current mass plus 2.494 21 × 10 15 kg kilograms ...
The choice of solar mass, M ☉, as the basic unit for planetary mass comes directly from the calculations used to determine planetary mass.In the most precise case, that of the Earth itself, the mass is known in terms of solar masses to twelve significant figures: the same mass, in terms of kilograms or other Earth-based units, is only known to five significant figures, which is less than a ...