Someone weighs 200 pounds on the surface of the Earth. If the radius of the Earth is 4000 miles, how much would that person weigh 1000 miles above the surface?

Question

Nathan Ashcroft · Accepted Answer

Person will weigh #128# pounds, #1000# miles above the surface.

The force of gravity acting on an object is its weight. As force of gravitation #F# is directly proportional to the two objects, here #m_o#, mass of object and #m_E#, mass of Earth and inversely proportional to the square of the distance between them. Constant of proportionality being #G#m the gravitational constant. In other words #F=Gxx(m_om_E)/r^2# and here #200=Gxx(m_om_E)/4000^2# i.e. #Gm_om_E=200xx4000^2# When the person (the object in our case) moves #1000# #km.# above Earth i.e. the distance between them increases to #5000# #km.#, the weight (i.e. the force) becomes #F=(Gm_om_E)/5000^2=(200xx4000^2)/5000^2# = #200xx16/25=128# pounds

Leo Abeyta · Answer

undefined To calculate the weight of the person 1000 miles above the surface of the Earth, you can use the formula for gravitational force:

$$ F = \frac{{G \times m_1 \times m_2}}{{r^2}} $$

Where:
- $ F $ is the gravitational force,
- $ G $ is the gravitational constant ($ 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 $),
- $ m_1 $ and $ m_2 $ are the masses of the two objects (in this case, the person and the Earth),
- $ r $ is the distance between the centers of the two objects.

Given:
- $ m_1 = 200 \, \text{lbs} $,
- $ r = 4000 + 1000 = 5000 \, \text{miles} $ (distance from the center of the Earth to the person's new position).

First, convert the weight of the person to mass:
$$ m_2 = \frac{{200 \, \text{lbs}}}{{g}} $$
Where $ g $ is the acceleration due to gravity on the surface of the Earth ($ 32 \, \text{ft/s}^2 $).

Then, calculate the gravitational force at the new distance using the given formula.

Adrian Ayala · Answer

undefined To calculate the weight of the person 1000 miles above the surface of the Earth, we can use the formula for gravitational force:

$$ F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} $$

Where:
- $ F $ is the gravitational force,
- $ G $ is the gravitational constant ($6.674 \times 10^{-11}$ N m$^2$/kg$^2$),
- $ m_1 $ and $ m_2 $ are the masses of the two objects (in this case, the person and the Earth),
- $ r $ is the distance between the centers of the two objects.

First, we find the gravitational force acting on the person on the surface of the Earth using their mass ($m_1$) and the Earth's mass, and the radius of the Earth ($r$).

$$ F_1 = \frac{{G \cdot m_1 \cdot M_{\text{Earth}}}}{{r_{\text{Earth}}^2}} $$

Then, we find the gravitational force acting on the person when they are 1000 miles above the surface of the Earth ($ r_{\text{Earth}} + 1000 $):

$$ F_2 = \frac{{G \cdot m_1 \cdot M_{\text{Earth}}}}{{(r_{\text{Earth}} + 1000)^2}} $$

Now, to find the weight of the person 1000 miles above the surface, we divide $ F_2 $ by the acceleration due to gravity on the surface of the Earth ($ g_{\text{Earth}} $):

$$ \text{Weight}_{\text{above}} = \frac{{F_2}}{{g_{\text{Earth}}}} $$

Using these formulas, we can calculate the weight of the person 1000 miles above the surface of the Earth.