Mass of the earth is increased by 2%, radius of the earth is increased by 4% find the percentage change in g.?

Question

Formula to be used:-

g=Gm1m2 divided by radius xradius.G=6.67x10 to the power of -11.Along with the solution can i get the answer???

Answer 1

the old g = GM/(R^2)

the new g = (1.02)/(1.04^2)(GM)/(R^2)

new g = 0.9430 old g

Answer 2

g = GM/R^2

Let M2 = new mass

R2 = new radius,

g2 = new value of g

M2 = M + 2% of M = 1.02 M

R2 = R+4% of R = 1.04 R

g2=G*1.02M/(1.04R)^2 = 1.02/1.04^2 GM/R^2

g2 = 1.02/1.04^2 g = 0.943 g

% change in g = 100 * (g2 - g)/g

= 100 *  (0.943 g - g)/g

= 100 * (0.943 - 1)

= -5.7

It is negative meaning that g decreases.

Ans: g decreases by 5.7 %

Answer 3

increase in mass will increase g by 2%. increase in radius will result in a 8% decrease in g, because r is in denominator and is squared.

total change is approximately -6%

These are all approximations. Based on:

1/(1+Δ) ≈ 1-Δ

(1+Δ)² ≈ 1+2Δ

edit stupid arith mistake

.

Answer 4

See: http://en.wikipedia.org/wiki/Earth%27s_g...

The formula you cite is that for the gravitational force, not the acceleration due to gravity, g

F = Gm1m2/r²

Anyway, let m1 be constant and m2 increase by 2% (factor 1.02), while r increases by 4% (factor 1.04). Since G is also constant, F', the new value of F will be F' = 1.02F/1.04² = 0.943F

Percentage change in F = 100(0.943F-F)/F = -5.7%

Answer 5

g =G m1 m2 \ d^2

g=(6.67x10 ^ -11 x 5.98 x 10^24 x 1) \ [(6.36 x 10^ 6)^2] = 9.86 m\s^2

Mass of the earth is increased by 2% so it's = 6,0996 x 10^24 kg

radius of the earth is increased by 4% so it's = 6614400 m

g =G m1 m2 \ d^2

g=(6.67x10 ^ -11 x  6,0996 x 10^24 x 1) \ [(6614400)^2] = 9,299 m\s^2

so the approximate result of the percentage change in g

= 100 * (0.943 - 1) = -5.7

so it is - 5.7% or g is decrased by 5.7%