Edmund Hillary had it easy

Classical Mechanics Level 2

Climbing to the top of Mount Everest is hard, but it's slightly easier than you might think as people weigh less as they climb to the top. Let $W_E$ be a person's weight on top of Mount Everest and $W_S$ be their weight at sea level. What is the value of $1-W_E/W_S$ ?

Details and assumptions

Assume the earth (other than Everest) is a sphere of mass $6 \times 10^{24}~\mbox{kg}$ and radius $6,370~\mbox{km}$ .
The top of Mount Everest is $8,848~\mbox{m}$ above the surface of the earth.

The answer is 0.0028.

2 solutions

Mashrur Fazla
Oct 14, 2013

Here,
G = $6.673\times 10^-11$ $Nm^2kg^-2$
M = $6\times 10^24$ kg
R = $6370$ km = $6370000$ m
h = $8848$ m

$g_{E}$ =( $6.673\times 10^-11 Nm^2kg^-2\times 6\times 10^24$ kg ) / $(6370000+8848)^2$ m
So, $g_{E}$ = $9.83983618$ $ms^-2$

Again, $g_{S}$ =( $6.673\times 10^-11 Nm^2kg^-2\times 6\times 10^24$ kg ) / $(6370000)^2$ m
So, $g_{S}$ = $9.867190446$ $ms^-2$

Now,1- $W_{E}/W_{S}$
=1- ( $m\times g_{E}$ )/ ( $m\times g_{S}$ )
=1- ( $9.83983618$ )/ ( $9.867190446$ )
=1- $0.9972277554$
= $2.8\times 10^-3$
SO,The answer is $\boxed{2.8\times 10^-3}$

Let x be your weight at sea level and r is the distance to the center of the earth

Your weight on the top of Everest will be (r / (r + height of Everest))^2 * x

Kshitij Khandelwal - 7 years, 8 months ago

yes you are also right. In this way
$W_{E}$ / $W_{S}$
= $(r / (r + h)^2 * x$ )/ $(r / r ^2 * x$ )
= $(6370000^2\times x$ )/ $(6378848^2\times x)$
= $0.9972277554$
Now $1-W_{E}/W_{E}$
= $1-0.9972277554$
= $2.8\times 10^-3$
SO.the answer is also $\boxed{2.8\times 10^-3}$

Mashrur Fazla - 7 years, 8 months ago

I used a calculator, got my answer as 0.0027 and I hate it now.

Abhijeeth Babu - 7 years, 7 months ago

You got $0.00277$ that means $0.0028$ .But if you give $0.00277$ as answer it will also be correct.

Mashrur Fazla - 7 years, 7 months ago

well done

Kartik Sharma - 7 years, 7 months ago

very lenthy way

rugved dhore - 7 years, 7 months ago

i round it to 3 x 10(sqr)-3 and it's wrong because of that :P

Hafizh Ahsan Permana - 7 years, 2 months ago

I posted the answer as 0.0027 by doing the same method.

Abin Das - 7 years, 7 months ago

is it correct? I posted 2.77E-3

Mashrur Fazla - 7 years, 7 months ago

Jamie Coombes
Oct 14, 2013

We begin by finding the ratio of the gravitational field strength on Everest to Sea level: $\frac{g_e}{g_s} = \frac{\frac{GM}{r_e^2}}{\frac{GM}{r_s^2}} = \frac{r_s^2}{r_e^2}$

Using Newton's second law, $F_e = ma_e$

$F_s = mg_s$

so the ratio of the two forces due to gravity are, $\frac{F_e}{F_s} = \frac{g_e}{g_s} = \frac{r_s^2}{r_e^2} = \frac{6370000^2}{6378848^2} = 0.997$

Finally, we subtract the ratio from one.

$1- \frac{F_e}{F_s} = 1 - 0.997 = 0.00277$