When a falling meteoroid is at a distance above the Earth's surface of 2.85 times the Earth's radius, what is itsacceleration due to the Earth's gravitation?Part 1 of 3 - ConceptualizeThe motion of the meteoroid, or its mass, makes no difference. The gravitational field, which is theacceleration, will be smaller than the surface value we are used to because of the greater distance.Part 2 of 3 - CategorizeThe Earth exerts its force on the meteoroid according to mg = GmM- /r so we see that acceleration ofgravity, g = GM- /r², follows an inverse-square law. At the surface, a distance of one Earth-radius (R-) fromthe center, it is 9.80 m/s². An altitude of 2.85 R- above the surface represents a distance of 3.85 R- from the center of the Earth, so theacceleration of gravity will be (3.85) = 14.82 times smaller.GME2GME2RE9.80 m/s2m/s².II||

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When a falling meteoroid is at a distance above the Earth's surface of 2.85 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter4: The Laws Of Motion

Section4.3: Newton's Second Law

Problem 4.3QQ: Respond to each statement, true or false: (a) No force of gravity acts on an astronaut in an...

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Central Force

Central force is the force that acts along the center of mass of one body to the center of mass of another body. Its direction from or away from a point on a body to another is called the center of the force. Such forces include gravitational force between two massive objects or electrostatic forces between two charged particles. So from atoms to planets, the central forces play an important role in shaping our daily lives. One of the remarkable features of central forces is that they are conservative, i.e. the total energy of a central force system can always be expressed as a gradient of a potential energy function.

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When a falling meteoroid is at a distance above the Earth's surface of 2.85 times the Earth's radius, what is its
acceleration due to the Earth's gravitation?
Part 1 of 3 - Conceptualize
The motion of the meteoroid, or its mass, makes no difference. The gravitational field, which is the
acceleration, will be smaller than the surface value we are used to because of the greater distance.
Part 2 of 3 - Categorize
The Earth exerts its force on the meteoroid according to mg = GmM- /r so we see that acceleration of
gravity, g = GM- /r², follows an inverse-square law. At the surface, a distance of one Earth-radius (R-) from
the center, it is 9.80 m/s².

An altitude of 2.85 R- above the surface represents a distance of 3.85 R- from the center of the Earth, so the
acceleration of gravity will be (3.85) = 14.82 times smaller.
GME
2
GME
2
RE
9.80 m/s2
m/s².
II
||

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