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Question: Two small bodies of masses \(10\) kg and \(20\) kg are kept a distance \(1.0\) m apart and released....

Two small bodies of masses 1010 kg and 2020 kg are kept a distance 1.01.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.50.5 m.

Explanation

Solution

To find the speeds of the particles when the separation between two bodies decreases, we have to use the concept of conservation of momentum and energy. Also, the gravitational potential energy acting on two bodies is given by the gravitational potential times the mass of the body.

Formula used:
Gravitational Potential energy of a two particle system is given by
U=−Gm1m2rU = - \dfrac{{G{m_1}{m_2}}}{r}
Where, GG - universal gravitational constant and m1,m2{m_1},{m_2} - masses separated by the distance rr.

Complete step by step answer:
Two bodies are separated by the distance 1.01.0 m and released. These two particles are approaching each other. So, using conservation of momentum, we have
m1v1+m2v2=0{m_1}{v_1} + {m_2}{v_2} = 0
⇒10v1+20v2=0\Rightarrow 10{v_1} + 20{v_2} = 0
We get, v1=2v2−−−−−−−−−−(1){v_1} = 2{v_2} - - - - - - - - - - (1)
Here, v1,v2{v_1},{v_2} are the velocities of bodies and considering only magnitude.
The gravitational potential energy is given by
U=−Gm1m2r=−6.67×10−11×10×201.0U = - \dfrac{{G{m_1}{m_2}}}{r} = - \dfrac{{6.67 \times {{10}^{ - 11}} \times 10 \times 20}}{{1.0}}
⇒U=−13.34×10−9J−−−−−−−(2)\Rightarrow U = - 13.34 \times {10^{ - 9}}J - - - - - - - (2)

Now, We know that the total energy of the system is always constant.Using Conservation of energy, we get;
Gravitational potential total energy =−Gm1m2r2+12m1v12+12m2v22 = \dfrac{{ - G{m_1}{m_2}}}{{{r_2}}} + \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2
⇒−13.34×10−9J=−6.67×10−11×10×200.5+1210v12+1220v22\Rightarrow - 13.34 \times {10^{ - 9}}J = \dfrac{{ - 6.67 \times {{10}^{ - 11}} \times 10 \times 20}}{{0.5}} + \dfrac{1}{2}10v_1^2 + \dfrac{1}{2}20v_2^2
⇒−13.34×10−9J=−13.34×10−90.5+20v22+10v22\Rightarrow - 13.34 \times {10^{ - 9}}J = \dfrac{{ - 13.34 \times {{10}^{ - 9}}}}{{0.5}} + 20v_2^2 + 10v_2^2
Using equation (1)&(2)(1)\& (2) ,
−13.34×10−9J=−13.34×10−90.5+20v22+10v22- 13.34 \times {10^{ - 9}}J = \dfrac{{ - 13.34 \times {{10}^{ - 9}}}}{{0.5}} + 20v_2^2 + 10v_2^2
We get, v22=4.44×10−10v_2^2 = 4.44 \times {10^{ - 10}}
v2=2.1×10−5ms−1{v_2} = 2.1 \times {10^{ - 5}}m{s^{ - 1}}
And from equation (1)(1) , v1=4.2×10−5ms−1{v_1} = 4.2 \times {10^{ - 5}}m{s^{ - 1}}

Hence, the speeds of the two bodies are respectively v1=4.2×10−5ms−1{v_1} = 4.2 \times {10^{ - 5}}m{s^{ - 1}} and v2=2.1×10−5ms−1{v_2} = 2.1 \times {10^{ - 5}}m{s^{ - 1}}.

Note: The conservation of energy theorem should be applied for each case and the sum of total mechanical energy of a system is conserved. Bodies having different speeds possess different kinetic energy but have potential energy as their distance is same for both bodies. We must know the value of universal gravitational constant. Do not use r2{r^2} in the denominator of gravitational potential energy as it will become Newton's law of gravitation.