Calculate Pat's Weight at Earth's Double Radius

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To calculate Pat's weight at an altitude twice the Earth's radius, the gravitational force formula Fg = G(m1)(m2)/r^2 is used, where G is the gravitational constant. Pat's mass is 65.5 kg, and at double the Earth's radius, the gravitational acceleration decreases due to the increased distance from the Earth's center. The gravitational force at this altitude can be expressed as Fg = (65.5 kg)(g') where g' is the new gravitational acceleration. The correct calculation shows that Pat's weight at this altitude is approximately 71.4 N. Understanding how gravitational force changes with distance is key to solving this problem.
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Homework Statement


Pat's mass is 65.5 kg on Earth's surface. What would Pat's weight be at an altitude twice the radius of the Earth


Homework Equations


Fg=G(m1)(m2)/r^2
G = 6.67X 10^-11
F=ma

We are not given the numerical radius of the Earth

The Attempt at a Solution


I tried to manipulate the formula a bit:
(using the gravitational law)
Fg = (65.5)x/ 2^2

And that got me nowhere.
I also know the answer is 71.4N but how you get there is what I don't understand.
 
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Fg=G(m1)(m2)/r^2

This equation should work. What are you using for each variable? Look up data you need. It's probably in your textbook somewhere.
 
While hage's solution is valid, I think we know g=9.8 for the surface of the Earth. Therefore, you can look at the equation and see how this force would change as R increases. Think about what is constant and what is changing.
 
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