Answer to Question #106056 in Classical Mechanics for Josephine

Question

1. Ceres is the largest asteroid. Its mean radius is 476 km and its mass is 9.4×10^20 kg. Using this, and G = 6.67×10^−11 Nm^2 kg^−2, and approximating it as a sphere, compute the speed required to escape the gravity on the surface of Ceres.

Speed = ___ m.s^−1. (to two significant figures, don't use scientific notation)

2. Consider the mechanical energy of a body in geostationary orbit above the Earth's equator, at rGS = 42000.

Consider the mechanical energy of the same body on Earth at the South pole, at re = 6400 km. For this problem, we consider the Earth to be spherical. (Remember, the object at the equator is in orbit, the object at the Pole is not in orbit.)

G = 6.67×10^−11 Nm^2 kg^−2, and the mass of the Earth is M = 5.97×10^24 kg.

What is the difference in the mechanical energy per kilogram between the two?

E= ___ MJ.kg^−1 (to two significant figures, don't use scientific notation).

Answer 1

1) Now, applying conservation of energy,

$\frac{GMm}{r}=\frac{mv^2}{2}$

$v=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2(6.67\cdot10^{−11})(9.4\cdot10^{20})}{467000}}=520\frac{m}{s}$

2) Mechanical energy  per kilogram at the surface of the earth

$E_1=\frac{GM}{R}=\frac{(6.67\cdot10^{−11})(5.97\cdot10^{24})}{6400000}=62.2\frac{MJ}{kg}$

Mechanical energy per kg in geostationary orbit above the Earth's equator

$E_2=\frac{GM}{r_{GS}}+0.5v^2=\frac{2GM}{r_{GS}}=\\=\frac{2(6.67\cdot10^{−11})(5.97\cdot10^{24})}{42000000}=19.0\frac{MJ}{kg}$

$\Delta E=62.2-19.0=43\frac{MJ}{kg}$