Tuesday

March 31, 2015

March 31, 2015

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.... - The universal law of gravitation describes the force of gravity ...

physics - he universal law of gravitation describes the force of gravity between...

Grade 12 Physics - A rocket ship is moving away from the earth after having been...

Physics - A charge of 1nC is added to a spherical soap bubble with a radius of 3...

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(...

physics - A 7.50-kg turkey standing at the top of Ishpatina Ridge, at 693 m the ...

Physics - A beam of protons is accelerated through a potential difference of 0....

physics.... - Spherical planet A has mass m and radius r. Spherical planet B has...

Physics - The gravitational potential difference between the surface of a planet...