How Does Distance from Earth Affect Gravitational Weight?

[Rgonzales]

Homework Statement

For this problem, use ratios only to obtain the weight of a person at the following distances. Assume the person weighs 980N on the surface of Earth.

a) 128 000km above the surface of Earth
b) 4.5 times the distance from the centre of Earth
c) 745 400km from Earths Centre.

Homework Equations

$$F_{g1} r_1^2 = F_{g2} r_2^2$$

The Attempt at a Solution

I am confused of where numbers should go. I assumed $$F_{g1}$$ to be 980N,
$$r_1^2$$ to be the radius of the Earth of 6.38 x 10(to the power of 6), $$r_2^2$$ to be 128 000 km and $$F_{g2}$$ to be the missing variable. Yet, i did not manage to find the answer. Please help me out.

 

[lippyka]

a) 128 000km above the surface of Earth F_{g1} (6.38 x 10(to the power of 6)) = F_{g2} (128 000 + 6.38 x 10(to the power of 6))980N (6.38 x 10(to the power of 6)) = F_{g2} (134.38 x 10(to the power of 6))F_{g2} = 980N/ (134.38 x 10(to the power of 6)) F_{g2} = 7.32 x 10(-7) Nb) 4.5 times the distance from the centre of Earth F_{g1} (6.38 x 10(to the power of 6)) = F_{g2} (4.5 x 6.38 x 10(to the power of 6))980N (6.38 x 10(to the power of 6)) = F_{g2} (28.71 x 10(to the power of 6))F_{g2} = 980N/(28.71 x 10(to the power of 6)) F_{g2} = 34.2 x 10(-7) Nc) 745 400km from Earths Centre F_{g1} (6.38 x 10(to the power of 6)) = F_{g2} (745400 + 6.38 x 10(to the power of 6))980N (6.38 x 10(to the power of 6)) = F_{g2} (751.78 x 10(to the power of 6))F_{g2} = 980N / (751.78 x 10(to the power of 6)) F_{g2} = 1.30 x 10(-7) N

 

[m2003]

As a scientist, it is important to carefully analyze and understand the problem before attempting to solve it. In this case, we are looking for the weight of a person at different distances from the surface of the Earth, using ratios and the given weight of the person on the surface (980N).

First, let's define the variables we will be using:

F_{g1} = gravitational force at first distance (in this case, on the surface of Earth)
F_{g2} = gravitational force at second distance
r_1 = radius of Earth (6.38 x 10^6 m)
r_2 = distance from Earth's center

Now, we can use the gravitational force equation:

F_{g1} = F_{g2} (r_1 / r_2)^2

For part a) 128 000km above the surface of Earth, we can calculate the distance from Earth's center using the given radius of Earth (6.38 x 10^6 m) and the given distance (128 000 km = 128 000 000 m).

r_2 = r_1 + 128 000 000 = 6.38 x 10^6 + 1.28 x 10^8 = 1.34 x 10^8 m

Plugging this into the equation, we get:

980N = F_{g2} (6.38 x 10^6 / 1.34 x 10^8)^2

Solving for F_{g2}, we get:

F_{g2} = 980N * (1.34 x 10^8 / 6.38 x 10^6)^2 = 98N

Therefore, the weight of the person at 128 000km above the surface of Earth is 98N.

For part b) 4.5 times the distance from the center of Earth, we can use the same process to calculate the distance from Earth's center:

r_2 = 4.5 * r_1 = 4.5 * 6.38 x 10^6 = 2.87 x 10^7 m

Plugging this into the equation, we get:

980N = F_{g2} (6.38 x 10^6 / 2.87 x 10^7)^2