- #1
stallion
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I am trying to figure out why F=mg corresponds to question number two below.
First let's assume a object with mass 70kg and then calcutate the force of gravity in two different scenarios:
Determine the force of gravitational attraction between the Earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.37 x 106 m from Earth's center.
Crunching the numbers yields 688.1 N
#2
Determine the force of gravitational attraction between the Earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above Earth's surface. This would place the student a distance of 6.38 x 106 m from Earth's center.
This yields: 686 Newtons at a distance of 40,000 feet.
Now use the 2nd law
F=mg (70kg)(9.8m/s^2) equals 686 N
Why is the 2nd Law calculation the same as the calculation at 40,000 feet?
I realize that this is a very small difference (less than 1%) between the two
numbers.
Thanks
First let's assume a object with mass 70kg and then calcutate the force of gravity in two different scenarios:
Determine the force of gravitational attraction between the Earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.37 x 106 m from Earth's center.
Crunching the numbers yields 688.1 N
#2
Determine the force of gravitational attraction between the Earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above Earth's surface. This would place the student a distance of 6.38 x 106 m from Earth's center.
This yields: 686 Newtons at a distance of 40,000 feet.
Now use the 2nd law
F=mg (70kg)(9.8m/s^2) equals 686 N
Why is the 2nd Law calculation the same as the calculation at 40,000 feet?
I realize that this is a very small difference (less than 1%) between the two
numbers.
Thanks
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