Question

Two spherical asteroids have the same radius $R$. Asteroid$1$has mass $M$and asteroid $2$has mass $2M.$The two asteroids are released from rest with distance $10R$between their centers. What is the speed of each asteroid just before they collide?

Hint: You will need to use two conservation laws.

Short answer

The speed of each asteroid is$v_1=-1.0328\sqrt{(GM/R)}$and$v_2=0.5164\sqrt{(GM/R)}$

Step-by-step solution

Given information

Mass of asteroid$1=M$

Mass of asteroid$2=2M$

Radius of each asteroid$=R$

Initial distance between their centers$=10R$

Initial velocity of both asteroids; $u=0m/s$(because they are released from rest).

Explanation

From law of conservation of momentum, we can write that;

$m_1•u_1+m_2•u_2=m_1•v_1+m_2•v_2$

Plugging in the relevant values gives;

$$\begin{gathered} M\left(0\right)+2M\left(0\right)=M\left(v_1\right)+2M\left(v_2\right) \\ M\left(v_1\right)+2M\left(v_2\right)=0 \\ {v}_1=-2\left(v_2\right) \end{gathered}$$

From law of conservation of energy, we know that;

$G{m}_1•m_2\left(\right(1/d\text{'})-(1/d\left)\right)=½m_1•\left(v_1\right)²+½m_2•\left(v_2\right)²$

Where;

$$\begin{gathered} d\text{'}=2R \\ d\text{'}=2Rd=10R \end{gathered}$$

Thus;

$G\times M\times 2M\left(\right(1/2R)-(1/10R\left)\right)=½M(-2v_2)²+½\left(2M\right(v_2)²)$

$$\begin{gathered} 2GM²(2/5R)=½M\left(v_2\right)²(4+2) \\ 2GM²(2/5R)=3M\left(v_2\right)² \end{gathered}$$

This reduces to;

$$\begin{gathered} 2GM(2/5R)=3\left(v_2\right)² \\ 4GM/5R=3\left(v_2\right)² \\ \left(v_2\right)²=4GM/15R \\ {v}_2=\sqrt{(4GM/15R)} \\ {v}_2=0.5164\sqrt{(GM/R)} \end{gathered}$$

Thus;$v_1=-2(0.5164\sqrt{(GM/R}))$

$v_1=-1.0328\sqrt{(GM/R)}$