What is the gravitational acceleration on Earth's surface?

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SUMMARY

The gravitational acceleration on Earth's surface is calculated using the formula \( g = \frac{G M_{earth}}{r^2} \), where \( G = 6.67 \times 10^{-11} \, m^3/kg/s^2 \), \( M_{earth} = 6.0 \times 10^{24} \, kg \), and \( r = 6378 \, km \) (converted to meters). This results in a gravitational acceleration of approximately 9.81 m/s². The discussion also emphasizes the importance of the Universal Law of Gravitation, which is fundamental in understanding gravitational interactions.

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I have no clue. Can someone show me how to calulate this.

Calculate the gravitational force between the Sun and the Earth. Use the following numbers. G = 6.67 x 10^(-11) m^3/kg/s^2, Mass of the Sun (M_sun) = 2.0 x 10^30 kg, Mass of the Earth (M_earth) = 6.0 x 10^24 kg, Distance between the Sun and the Earth (r) = 1.5 x 10^8km.
 
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The magnitude of the gravitational force is given by:

<br /> F=\frac{G m_{earth}M_{sun}}{r^2}<br />

Just plug in the numbers.
 
Just as an added note, the equation that cyosis posted is called the "Universal law of gravitation" or "Newton's universal law of gravitation". Remember it and remember it well young padawan.

The general form is just without the sun and Earth subscripts, aka:

F = \frac{G M_1 M_2}{R^2}

Extra credit question for you: determine the gravitational acceleration of Earth while standing on the surface, if the radius is ~6378 km.
 

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