Question

Two hypothetical planets of masses m1 and m2 and radii r1 and r2, respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. Note: Both the energy and momentum of the isolated two planet system are constant.(a) When their center-to-center separation is d, find expressions for the speed of each planet and for their relative speed. (Use any variable or symbol stated above along with the following as necessary: G, the gravitational constant.)v1 =___________v2 =__________vr =_________(b) Find the kinetic energy of each planet just before they collide, taking m1 = 2.60 1024 kg, m2 = 7.00 1024 kg, r1 = 2.80 106 m, and r2 = 5.80 106 m.K1 = ________JK2 = ________-J

Answer 1

Step-by-step explanation:

a )

When they are at infinite separation , energy of the system is zero

When they are at d separation

potential energy = - G m₁m₂ / d

So there is reduction in potential energy which will increase their kinetic energy by the same amount so that total energy becomes zero again .

Let their velocity becomes v₁ and v₂ respectively

according to c

law of conservation of momentum

m₁v₁ = m₂v₂ ( in magnitude )

Their total kinetic energy

= 1/2 m₁v₁² + 1/2 m₂v₂² = G m₁m₂ / d

= 1/2 m₁v₁² + 1/2 m₂(m₁v₁ / m₂)² = G m₁m₂ / d

= 1/2 m₁v₁² + 1/2 m₁²v₁² / m₂ = G m₁m₂ / d

1/2 m₁v₁²( 1 + m₁ / m₂ ) = G m₁m₂ / d

v₁ =
`\sqrt{(2Gm_2^2)/(d(m_1+m_2)) }`

Similarly

v₂ =
`\sqrt{(2Gm_1^2)/(d(m_1+m_2)) }`

Relative velocity = v₁ +v₂

=
`\sqrt{(2Gm_2^2)/(d(m_1+m_2)) } + \sqrt{ (2Gm_1^2)/(d(m_1+m_2)) }`

Putting the values in the expression of v₁

we get

v₁ = 7.9 x 10⁷ m / s

v₂ = 1.09 x 10⁷

Kinetic energy of m₁

= .5 x 2.6 x 10²⁴ x (7.9 x 10⁷)²

= 81.13 x 10³⁸ J