A dwarf planet with an average orbital distance of 4.14 × 108 kilometers is kept

Question

A dwarf planet with an average orbital distance of 4.14 × 108 kilometers is kept in its orbit by 6.94 × 10 17 Newton force exerted on it by the Sun. MSun = 1.998 1030 kg. MEarth = 5.97 × 1024 kg. What is the mass of that planet? How many Earth masses is that equivalent to? kilograms (use scientific notation and round the coefficient to two decimals) (use scientific notation and round the coefficient to two decimals) QUESTION 5 An asteroid with mass of 2.59 x 1020 kilograms or 4.34 × 10-5 MEarth is kept in its orbit by 2.75x 1017 Newton force exerted on it by the Sun. MSun = 1.99 x 1030 kg. 1 AU = 1.5x 108 km What is the radius of that asteroid's orbit? kilometers (use scientific notation and round the coefficient to two decimals) What is the radius of that asteroid's orbit in AU? AU (round your answer to two decimals).

Explanation / Answer

Gravitational force exerted by sun on the planet

F = GM­sunMplanet/(rorbit)2

6.94 x 1017 N = 6.67 x 10-11 x 1.99 x 1030 x MPlanet / (4.14 x 108 )2

MPlanet = 6.94 x 1017 x (4.14 x 108)2 /6.67 x 10-11 x 1.99 x 1030

MPlanet = 8.96 x 1014 Kg

Mplanet/Mearth = 8.96 x 1014/5.97 x1024

Mplanet =(1.50 x 10-10) MEarth

Mass of the planet is 8.96 x 1014 Kg and is equal to 1.50 x 10-10 earth masses

5)

Gravitational force exerted by sun on the asteroid

F = GM­sunMasteroid/(rorbit)2

2.75 x 1017 N = 6.67 x 10-11 x 1.99 x 1030 x 2.59 x 1020/ (rorbit )2

rorbit2 = 6.67 x 10-11 x 1.99 x 1030 x 2.59 x 1020/2.75 x 1017

rorbit = 3.54 x 1011 m

rorbit = 3.54 x 108 Km

rorbit = 3.54 x 108 / 1.5 x 108 = 2.36 AU

Radius of the orbit is 3.54 x 108 Km and 2.36 AU

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