Question

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

(a) Calculate the kinetic energy that the earth has because of its rotation about its own axis. For comparison, the total energy used in the United States in one year is about 9.33 multiplied by 109 J.

(b) Calculate the kinetic energy that the earth has because of its motion around the sun.

Answer 1

a. 7.43 × 10³⁴ J b. 3.51 × 10³⁸ J

Step-by-step 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

Answer 2

Explanationhe rotational kinetic energy is

`K_(r) =(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 )(
`((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

.