Question

Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2% keeping all other quantities same, then [BHU 1994; JIPMER 2000]
A) g decreases by 2% and K decreases by 4% B) g decreases by 4% and K increases by 2% C) g increases by 4% and K increases by 4% D) g decreases by 4% and K increases by 4%

Answer 1

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Secondary SchoolPhysics 5+3 pts


The acceleration due to gravity on the Earth’s surface is ‘g’. If the acceleration due to gravity on the surface of another planet, whose mass is four times the earth’s mass and radius is twice the earth’s radius is ng, find the value of n.
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Answers

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tiwaavi
Tiwaavi ★ Brainly Teacher ★
Hello Dear.

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For the Earth,

Let the Mass of the Earth be m and the radius of the earth be r.
Acceleration due to the gravity on Earth = g

Using the Formula,

g = Gm/r² --------------------------eq(i)

Where G = Gravitation constant
= 6.67 × 10⁻¹¹ Nm²/kg²

For the Another Planet,

Mass = 4 × Mass of the earth
Mass = 4 × m

Radius = 2 ×Radius of the Earth
= 2 × r

Acceleration due to the Gravity = ng

Now again using the formula,

Acceleration due to gravity = G × Mass/Square of the radius


⇒ ng = G × 4m/(2r)²
ng = G × 4m/4r²
ng = G × m/r² -------------------------------eq(ii)

Putting the Value of g from eq(i) in the above equation,We get,

n × (G × m/r²) = G × m/r²

⇒ n = 1

Thus, the value of n is 1.

Answer 2

Answer:

g = (GM / R2)

KE = (L2 / 2I)

for earth, g = acceleration due to gravity

K = KE(rotational)

Radius is decrease by 2%

if G & M remains constant then g ∝ (1 / R2)

(g2 / g1) = (R1 / R2)2

g2 = g1[R1 / (0.98R1)]2

g2 = 1.04g i.e. g increases by 4%

KE = (L2 / 2I) = [L2 / {2 × (2/5)MR2}] = [(5L2) / (4MR2)]

i.e. KE ∝ (1/R2)

KE is also increase by 4%