gravitation with the Earth as m2 and the radius of the Earth for the distance F = GmME RE2. Now all we need to do is substitute values and calculate. We already said m = 51.0 kg, and we know G = 6.67 ✕ 10−11 N · m2/kg2. The Earth is not a perfect sphere, but, its average radius is RE = 6.37 ✕ 106 m. The mass of the Earth is ME = 5.97 ✕ 10−24 kg. We can then substitute these values in the following formula. (Enter your answer in N.) F = (6.67 ✕ 10−11 N · m2/kg2)(51.0 kg)(5.97 ✕ 1024 kg) (6.37 ✕ 106 m)2(A) = _______ N Now let's compare this result to the person's weight (in N) found by multiplying the person's mass by g (or, that is, w = mg) where g = 9.80 m/s2. w = (51.0 kg)(9.80 m/s2) =(b) __________________ N You should have found that these two methods give about the same result! This tells us that g, the acceleration due to gravity, is actually determined by mass and radius of the Earth—or the Moon, or Mars, or whatever planet we happen to be standing on.
gravitation with the Earth as m2 and the radius of the Earth for the distance F = GmME RE2. Now all we need to do is substitute values and calculate. We already said m = 51.0 kg, and we know G = 6.67 ✕ 10−11 N · m2/kg2. The Earth is not a perfect sphere, but, its average radius is RE = 6.37 ✕ 106 m. The mass of the Earth is ME = 5.97 ✕ 10−24 kg. We can then substitute these values in the following formula. (Enter your answer in N.) F = (6.67 ✕ 10−11 N · m2/kg2)(51.0 kg)(5.97 ✕ 1024 kg) (6.37 ✕ 106 m)2(A) = _______ N Now let's compare this result to the person's weight (in N) found by multiplying the person's mass by g (or, that is, w = mg) where g = 9.80 m/s2. w = (51.0 kg)(9.80 m/s2) =(b) __________________ N You should have found that these two methods give about the same result! This tells us that g, the acceleration due to gravity, is actually determined by mass and radius of the Earth—or the Moon, or Mars, or whatever planet we happen to be standing on.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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So let's consider a person with a mass of 51.0 kg standing on the Earth. To find the gravitational force on the person, we'll again use Newton's law of universal gravitation with the Earth as
m2
and the radius of the Earth for the distanceF =
.
| GmME |
| RE2 |
Now all we need to do is substitute values and calculate. We already said
m = 51.0 kg,
and we know
G = 6.67 ✕ 10−11 N · m2/kg2.
The Earth is not a perfect sphere, but, its average radius is
RE = 6.37 ✕ 106 m.
The mass of the Earth is
ME = 5.97 ✕ 10−24 kg.
We can then substitute these values in the following formula. (Enter your answer in N.)F =
(A) = _______ N
| (6.67 ✕ 10−11 N · m2/kg2)(51.0 kg)(5.97 ✕ 1024 kg) |
| (6.37 ✕ 106 m)2 |
Now let's compare this result to the person's weight (in N) found by multiplying the person's mass by g (or, that is,
w = mg)
where
g = 9.80 m/s2.
w = (51.0 kg)(9.80 m/s2) =(b) __________________ N
You should have found that these two methods give about the same result! This tells us that g, the acceleration due to gravity, is actually determined by mass and radius of the Earth—or the Moon, or Mars, or whatever planet we happen to be standing on.
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