Thursday

November 27, 2014

November 27, 2014

Posted by **Larry** on Monday, January 21, 2013 at 10:51am.

If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is the mountain?

F= g ( m1 x m2)/ radius^2

f= gravitational force

g= 6.67x10^-11

m= mass

d= radius^2

earth mass= 5.98x10^24

earth radius = 6.38x10^6

- Physics -
**Steve**, Monday, January 21, 2013 at 11:29amYou know that at sea level

GM/r^2 = 9.8

Since G and M don't change, at radius R at the mountain top,

GM/R^2 = 9.772

so, R^2/r^2 = 9.8/9.772 = 1.0028653

R = 1.0014316r

R-r = 0.0014316r = 9134m

That seems a bit high. Better check my math.

- Physics -
**drwls**, Monday, January 21, 2013 at 11:33amThe actual height of Everest is 8848 m. Your number is close, and you used the right method. The difference is probably due to the use of too few significant figures for G(or GM/r^2).

**Answer this Question**

**Related Questions**

physics - How would one plug the gravitational constant into an equation? To be ...

Trigonometry - Okay, I have been given a trigonometric equation to solve (sin^2(...

System of equations - 3x =2y = 17 2x-y =9 and 3x+y = 3 x+2y = -4 How do I do ...

Integrating Factors - I've been working on this hw problem for a while now, but ...

Algebra ?? - What would be the missing piece of information given the following ...

Math: Differential Equations - I've been stuck on this math hw problem for a ...

Parabola - A parabola has been drawn on a graph but there is information missing...

maths equations - could you please help me on this equation r = sx2/2t im really...

solution of an equation - can someone please give me an example of a solution of...

algebra help please - Business and finance. The cost of producing a number of ...