Newton's Theory of Gravity and impact speed

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SUMMARY

The discussion focuses on calculating the impact speed of a 1.0 kg object released from a height of 500 km above Earth, utilizing Newton's Theory of Gravity. The correct formula for impact speed, considering gravitational potential energy and ignoring air resistance, is derived from energy conservation principles. The initial attempt at calculation was flawed due to dimensional inconsistencies and incorrect variable usage. Clarification is needed regarding the conditions of a flat Earth scenario for accurate impact speed determination.

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Kalie
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A 1.0 kg object is released from rest 500 km above the earth.
A. What is its impact speed as it hits the ground? Ignore air resistance
B. What would the impact speed be if the Earth were flat?
I don't know how to start this.
I said that v=sqrt(2G*M_e*m/(R_e+300))= 1.33*10^9 but that is incorrect...I'm probably being stupid but please help
 
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Kalie said:
A 1.0 kg object is released from rest 500 km above the earth.
A. What is its impact speed as it hits the ground? Ignore air resistance
B. What would the impact speed be if the Earth were flat?
I don't know how to start this.
I said that v=sqrt(2G*M_e*m/(R_e+300))= 1.33*10^9 but that is incorrect...I'm probably being stupid but please help
You seem to recognize that this problem has something to do with energy conservation. With the reference of zero potential at infinity the object has potential energy at altitude, but also when it hits the ground. You have several dimensional inconsistencies in your equation. What is 300? Why are there both M_e and m on the right and no m on the left.?

I have no idea what part C is about. Flat how? A cube? A huge plane of mass? I have no idea.
 

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