Question

A weight watcher who normally weighs 400 N stands on top of a very tall ladder so she is one Earth radius above Earth's surface (i.e. twice her normal distance from Earth's center). How much is her weight there? Group of answer choices

Answer 1

Answer:

100 N

Explanation:

Let us first find the mass of the person when they are on the earth's surface

$F=G\frac{Mm}{r^{2} }$

Now let us see how F changes if the distance between two masses is doubled

$F_{1} = G\frac{Mm}{(2r)^{2} } = \frac{1}{4} *G\frac{Mm}{r^{2} }$

So when the distance of separation is doubled, the force of gravity between the two masses decreases by a factor of four

Therefore her new weight = 400 N/4 = 100 N