Calculating Gravitational Force between Neptune and the Sun

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SUMMARY

The discussion centers on calculating the gravitational force between Neptune and the Sun using the formula derived from Newton's law of universal gravitation. The mass of the Sun is established as 332,800 times that of Earth, while Neptune's mass is 17 times that of Earth. The average orbital radius of Neptune is specified as 4503.99 x 109 m. The gravitational constant used is 6.67 x 10-11 N m2/kg2. The participant's initial calculation yielded 6.67 x 1020 N, which was incorrect, indicating a need for a review of the formula application or arithmetic.

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  • Understanding of Newton's law of universal gravitation
  • Familiarity with scientific notation and unit conversions
  • Basic knowledge of mass and gravitational constant values
  • Ability to perform algebraic calculations involving large numbers
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  • Practice calculations involving gravitational force with different celestial bodies
  • Explore the implications of gravitational force on planetary orbits
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Homework Statement



The mass of the sun is 332800 times that of the mass of Earth. The mass of neptune is 17 times that of the mass of the Earth. If the average orbital radius (distance from the sun) of neptune is 4503.99 x109 m, calculate the average gravitational force that neptune experiences due to the sun. Express your answer in units of 1021 N.
(The gravitational constant is 6.67 x 10^-11 N m2/kg2 and the mass of Earth is 5.98 x10^24 kg.)


The Attempt at a Solution



ok so i set up the problem like this..

(6.67 x 10^-11)(1.0166 x 10^26)(1.99 x 10^36) all over
(4503.99 x 10^9)^2

= 6.67 x 10^20 - which is wrong; I've used 1/3 tries on lon-capa and i kinda want to know what I am doing wrong here..

** to get the mass of the sun and neptune i just times them by the mass of Earth (5.98 x10^24)(17) for neptune and (5.98 x10^24)(332800) for the sun
 
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the answer is given as

something x10^21 N
 
solved it.. thanks for no help
 

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