Question

Assume that the earth is a uniform sphere and that its path around the sun is circular.

Answer 1

Explanationhe rotational kinetic energy is

$K_{r} =\frac{1}{2} Iω^{2}$

The moment of inertia  I  for a sphere is  ( 2 / 5 ) m r ^2

. Substituting this in the equation yields

Kr=1/2( ( 2 / 5 ) m r ^2 )($(\frac{v}{r})^{2}$

1/5mv^2

1/5*5.97 × 10 ^24 *(2$\pi$*6.38*10^6/86400)^2

2.57 × 10 ^29 J

b. kinetic energy of the sun

K.E=1/2*mv^2

the distance from the earth to the sun is given as

.

Answer 2

Answer:

a. 7.43 × 10³⁴ J b. 3.51 × 10³⁸ J

Explanation:

a. The gravitational force of attraction of a body on the surface of the earth equals the centripetal force on it due to the earth.

So, GMm/R² = mRω²

ω = √(GM/R³) where ω = angular speed of the earth. M = mass of earth = 5.972 × 10²⁴ kg, R = radius of earth = 6.4 × 10⁶ m and G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

The rotational kinetic energy of earth K.E = 1/2Iω² where I = rotational inertia = 2/5MR²

K.E = 1/2Iω²

= 1/2 × 2/5MR² × GM/R³

= GM²/5R

= 6.67 × 10⁻¹¹ Nm²/kg² × (5.972 × 10²⁴ kg)² /(6.4 × 10⁶ m × 5)

= 7.43 × 10³⁴ J

b. Similarly, the rotational kinetic energy of the earth around the sun is

K.E = GM²/5R where M = mass of sun = 1.989 × 10³⁰ kg and R = distance of earth from sun = 1.5047 × 10¹¹ m

K.E = GM²/5R

= 6.67 × 10⁻¹¹ Nm²/kg² × (1.989 × 10³⁰ kg)² / (1.5047 × 10¹¹ m × 5)

= 3.5073 × 10³⁸ J ≅ 3.51 × 10³⁸ J