IMAGINE- a building 6400km. high. On the ground floor, a person weighs 175lbs when he steps on a spring scale how much would the man weigh on the same scale if he were standing at the top floor? HINT: Note that 6400km is also the radius of the earth so that the top floor is 2Re from the earths center. Think about what this means for Fgrav before you start plugging in numbers or making conversions
The weight is equal ‘mg”
On the Earth surface acceleration due to gravity is
mg=GmM/R²,
g=G•M•/R²,
where
the gravitational constant G =6.67•10^-11 N•m²/kg²,
Earth’s mass is M = 5.97•10^24 kg,
Earth’s radius is R = 6.378•10^6 m.
g=9.8 m/s²
Let's find “g” at the height “h” . Distance between the mas “m” (the person) and the center of the Earth is 2R =>
m•g1= G•m•M/(2R)²
g1= G•M/4R²=g/4
In the "high" building mg1= mg/4
which ones do i not plug in? since every letter has something that represents it. Im confuesed.
To determine how much the person would weigh on the spring scale at the top floor of a building 6400km high, we need to consider the change in gravitational force experienced at different heights.
According to Newton's Law of Universal Gravitation, the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In this case, the two objects of interest are the person and the Earth.
Let's start by examining the relationship between the force of gravity and the distance from the center of the Earth. Considering that the ground floor of the building is at the same level as the Earth's surface, the radius of the Earth (Re) can be used as a reference point.
At the ground floor, the person weighs 175 lbs, which is the force of gravity acting on them. This force is denoted as F_grav_ground.
At the top floor, which is 2Re from the Earth's center (6400 km), we can assume that the person is at a distance of 3Re from the center.
Now, according to the inverse square relationship, the force of gravity is inversely proportional to the square of the distance. So, the force of gravity at the top floor, F_grav_top, can be expressed as:
F_grav_top = (F_grav_ground) * (d_grav_ground / d_grav_top)^2
Here, d_grav_ground refers to the distance from the center of the Earth to the ground floor (which is Re), and d_grav_top refers to the distance from the center of the Earth to the top floor (which is 3Re).
Plugging in the values:
F_grav_top = (175 lbs) * (Re / 3Re)^2
Simplifying further:
F_grav_top = (175 lbs) * (1/9)
F_grav_top = 19.44 lbs
Therefore, the man would weigh approximately 19.44 lbs on the same spring scale if he were standing at the top floor of a building 6400km high.