- #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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