Calculate Pat's Weight at Earth's Double Radius

In summary, to calculate Pat's weight at an altitude twice the radius of the Earth, we can use the formula Fg=G(m1)(m2)/r^2, where G=6.67x10^-11, m1 is Pat's mass (65.5 kg), and r is the distance from the center of the Earth. Since we are not given the numerical radius of the Earth, we can use the surface gravity of Earth, g=9.8, as a constant. This gives us an answer of 71.4N for Pat's weight at twice the Earth's radius.
  • #1
susan__t
20
0

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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  • #2
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.
 
  • #3
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.
 

1. How do you calculate Pat's weight at Earth's double radius?

To calculate Pat's weight at Earth's double radius, we would use the formula for gravitational force: F = (G*m1*m2)/d^2, where G is the gravitational constant, m1 is Pat's mass, m2 is the mass of the Earth, and d is the distance between Pat and the Earth's center. We would substitute in the values for G, m1, and m2, and then multiply the result by 2 since the distance is now double the original. The final result would be Pat's weight at Earth's double radius.

2. What is the gravitational constant (G)?

The gravitational constant (G) is a fundamental constant that represents the strength of the gravitational force between two objects. It has a value of approximately 6.67 x 10^-11 N*m^2/kg^2 in the SI unit system.

3. Is Pat's weight at Earth's double radius the same as his weight on the Moon?

No, Pat's weight at Earth's double radius would still be different from his weight on the Moon. This is because the Moon has a different mass and a different distance from Pat compared to Earth's double radius. Therefore, the gravitational force between Pat and the Moon would be different.

4. How does Earth's double radius affect Pat's weight?

As the distance between Pat and the Earth's center increases, the gravitational force between them decreases. This means that Pat's weight at Earth's double radius would be less than his weight at the original distance from Earth's center.

5. Can Pat's weight at Earth's double radius be negative?

No, Pat's weight at Earth's double radius cannot be negative. The formula for gravitational force only produces positive values, so even if the distance is doubled, the result would still be a positive number representing Pat's weight at that distance.

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