Question

At a place, value of 'g' is less by 1% than its value on the surface of the earth ( radius of earth = 6400 km) . the place is

a) 64 km below the surface of earth

b) 64 km above the surface of the earth

c) 30 km above the surface of the earth

d) 32 km below the surface of the earth

Answer 1

@mukushla ..

Answer 2

@ujjwal

Answer 3

g = GM/R^2

Answer 4

g = 9.8 - (9.8*1%)=9.8 - 0.098=9.702 m/s^2

Answer 5

I had this formula. It works for small ranges..\[g'=g(1-\frac{d}{R})\]

Answer 6

9.702 = 9.8(1-d/6400)

Answer 7

? like this ..@ujjwal

Answer 8

yep!

Answer 9

so .. i am getting 65306 meter ( approx. )

65306 m = 64 km (approx.)

Answer 10

and the direction is +ve hence it will be b) 64 km above the surface of earth .. ?

Answer 11

That is for below the surface of earth! Above the surface of earth the formula is:\[g'=g(1-\frac{2h}{R})\]

Answer 12

oh .. but i had to calculate whether it is above or below .. the surface of the earth

it is not given then which formula should i use

Answer 13

The answer should be 64 km below the surface of earth.
or, 32 km above it.( But we don't have that in the option.)

So, the answer is 64 km below the surface of earth.

Answer 14

but the book says 2nd option ..

Answer 15

That is impossible..
For above the surface of earth you can directly use the relation:
\[g=\frac{GM}{(R+h)^2}\]you have\[9.8=\frac{GM}{R^2}\]

Answer 16

im getting 32 km above the surface of the earth ??!!

Answer 17

yeah, the answer is both 64 km below the surface of earth and 32 km above the surface of earth.. The value of g is same at both places. But since we have only 64 km below the surface of earth in the option, the correct answer is (a)

Answer 18

http://www.wolframalpha.com/input/?i=At+a+place%2C+value+of+%27g%27+is+less+by+1%25+than+its+value+on+the+surface+of+the+earth+%28+radius+of+earth+%3D+6400+km%29+.+the+place+is+

Answer 19

\[F_g=\frac{GM}{r^2}\]\[G=6.67x10^{-11}\frac{m^3}{kg\space s^2}\]\[M_{earth}=5.972x10^{24}kg\]The question states that g is 1% lower at some point than it is at the surface of the earth...so the location of the point MUST be above the surface (since Fg is inversely proportional to the radius). That effectively eliminates answers a and d.

To figure out whether the answer is b or c, first calculate the force of gravity at the earth's surface using the radius given:
\[ Fg_{s}=\frac{(6.67x10^{-11}\frac{m^3}{kg\space s^2})(5.972x10^{24}kg)}{(6.4x10^6m)^2}=9.73m/s^2\]That's a little less than 9.8 m/s^2 but it should be...the earth's radius (depending on where you're at) is actually somewhere around 6356km-6378km.

So...1% below that would be:\[ (9.73m/s^2)(0.99)=9.63m/s^2\]
At 64km above the surface: \[Fg_{s+64km}=\frac{(6.67x10^{-11}\frac{m^3}{kg\space s^2})(5.972x10^{24}kg)}{(6.464x10^6m)^2}=9.53m/s^2\]
At 30km above the surface: \[ Fg_{s+30km}=\frac{(6.67x10^{-11}\frac{m^3}{kg\space s^2})(5.972x10^{24}kg)}{(6.430x10^6m)^2}=9.63m/s^2\]
Therefore, the only correct answer is c.

Answer 20

@Shane_B The value of g below the surface of earth is less than that on surface. Please view this for better explanation. And click "Show Explanation" in that page.. http://www.my-lyceum.net/MyLyceum/content/General%20Knowledge/Home.aspx?ItemIndex=5.1&CatalogName=general%20knowledge&CategoryName=Acceleration%20due%20to%20Gravity&TargetURL=Home.aspx