- #1

Angela G

- 65

- 19

- Homework Statement
- HI! I wonder if you could please help me with this excercise. It is from a course about planets.

A planet of mass M and radius R initially an angular velocity of w and no obliquity. It is impacted nearly tangentially at its north pole by a body of mass m≪ M, whose relative velocity before the encounter is small compared to the escape speed from the planet’s surface

Derive an expression for the planet’s angular velocity and obliquity after the impact in terms of M, m,R, ##\omega##,

- Relevant Equations
- $$ I = \frac{2}{5}MR^2$$

$$ L = r x p$$

$$ L = Iw $$

HI!

I tried to solve this exercise, by assuming that it is an inelastic collision, the planet is spherical, and that the rotation axis is parallel to the z-axis, see the figure attached. (1) before the collision, (2) after the collision.I started by assuming angular momentum conservation, which means that ## L_i = L_f## where 'i' refers to the initial and 'f' is the final.

The initial angular momentum ## L_i = L_o +L_p ## ( is the sum of the angular momentum of the object and the planet). The angular momentum of the planet is given by ## L_p = I * \omega ## where ## \omega## is the angular velocity before the collision, and ##I## is the moment of inertia ## I = \frac{2}{5} M*R^2## . The angular momentum of the object is given by ## L_o = r \times p = r sin(\theta)mv##. For the initial values, we set that ##\theta = 90## since the object hits the planet tangentially. So we have that ## L_o = Rmv##. For the final angular momentum, we have that ## L_f = I' \omega '##. We equate both side and get

$$ \frac{2}{5} M*R^2 * \omega + Rmv = I' \omega ' \Longrightarrow \frac{2}{5} M*R^2 * \omega + Rmv = \frac{2}{5}(M+m)R^2 \omega ' $$

we solve for ## \omega '## and we get

$$ \frac{5}{2(M+m)R^2}\left(\frac{2}{5} M*R^2 * \omega + Rmv \right)= \omega ' $$

But the problem is that I have still a term with the velocity of the object, which I should not get. I wonder if someone can please help me with this and if I can also solve this with forces.

Best regards

Angela

I tried to solve this exercise, by assuming that it is an inelastic collision, the planet is spherical, and that the rotation axis is parallel to the z-axis, see the figure attached. (1) before the collision, (2) after the collision.I started by assuming angular momentum conservation, which means that ## L_i = L_f## where 'i' refers to the initial and 'f' is the final.

The initial angular momentum ## L_i = L_o +L_p ## ( is the sum of the angular momentum of the object and the planet). The angular momentum of the planet is given by ## L_p = I * \omega ## where ## \omega## is the angular velocity before the collision, and ##I## is the moment of inertia ## I = \frac{2}{5} M*R^2## . The angular momentum of the object is given by ## L_o = r \times p = r sin(\theta)mv##. For the initial values, we set that ##\theta = 90## since the object hits the planet tangentially. So we have that ## L_o = Rmv##. For the final angular momentum, we have that ## L_f = I' \omega '##. We equate both side and get

$$ \frac{2}{5} M*R^2 * \omega + Rmv = I' \omega ' \Longrightarrow \frac{2}{5} M*R^2 * \omega + Rmv = \frac{2}{5}(M+m)R^2 \omega ' $$

we solve for ## \omega '## and we get

$$ \frac{5}{2(M+m)R^2}\left(\frac{2}{5} M*R^2 * \omega + Rmv \right)= \omega ' $$

But the problem is that I have still a term with the velocity of the object, which I should not get. I wonder if someone can please help me with this and if I can also solve this with forces.

Best regards

Angela