Maximizing and Minimizing Distance of a Bullet Fired from a Satellite

[Tanya Sharma]

Homework Statement

An Earth satellite is revolving in a circular orbit of radius 'a' with velocity 'v₀'. A gun is in the satellite and is aimed directly towards the earth.A bullet is fired from the gun with muzzle velocity v₀/2.Neglecting resistance offered by cosmic dust and recoil of gun,calculate maximum and minimum distance of bullet from the center of Earth during its subsequent motion.

Homework Equations

The Attempt at a Solution

Orbital speed of satellite is \sqrt{\frac{GM}{a}}

Initial velocity of the bullet v_{i} = \sqrt{{v_o}^2+(\frac{v_0}{2})^2} = \frac{\sqrt{5}v_{0}}{2}

Let P be the point at which bullet is fired and Q be point where distance is maximum/minimum.

Applying conservation of angular momentum at P and Q

mv_{i}a=mvr

or , v = \frac{v_{i}a}{r} = \frac{\sqrt{5}}{2}\frac{av_0}{r}

Applying conservation of mechanical energy at P and Q

\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}

Solving the equations , I get 3r^2-8ar+5a^2 = 0 which gives r =5/3a and a .

The answer i am getting is incorrect .

The correct answer given is 2a and 2a/3 .

I would be grateful if somebody could help me with the problem.

 

[gneill]

Homework Statement

An Earth satellite is revolving in a circular orbit of radius 'a' with velocity 'v₀'. A gun is in the satellite and is aimed directly towards the earth.A bullet is fired from the gun with muzzle velocity v₀/2.Neglecting resistance offered by cosmic dust and recoil of gun,calculate maximum and minimum distance of bullet from the center of Earth during its subsequent motion.

Homework Equations

The Attempt at a Solution

Orbital speed of satellite is \sqrt{\frac{GM}{a}}

Initial velocity of the bullet v_{i} = \sqrt{{v_o}^2+(\frac{v_0}{2})^2} = \frac{\sqrt{5}v_{0}}{2}

Let P be the point at which bullet is fired and Q be point where distance is maximum/minimum.

Applying conservation of angular momentum at P and Q

mv_{i}a=mvr

or , v = \frac{v_{i}a}{r} = \frac{\sqrt{5}}{2}\frac{av_0}{r}

Applying conservation of mechanical energy at P and Q

\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}

Solving the equations , I get 3r^2-8ar+ {\color{red}{5a^2}} = 0 which gives r =5/3a and a .

Check your algebra for that last term in your quadratic. Otherwise you've done fine up to that point.

 

[Tanya Sharma]

Hi gneill...

Sorry...I couldn't find any algebraic error .I redid the calculations . Maybe I am committing the same mistake again .

I keep on getting r=5a/3 and a .

 

[gneill]

Hi gneill...

Sorry...I couldn't find any algebraic error .I redid the calculations . Maybe I am committing the same mistake again .

I keep on getting r=5a/3 and a .

Can't fix what we can't see...

Can you elaborate your derivation of the quadratic a bit?

 

[Tanya Sharma]

\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}

\frac{1}{2}m\frac{5}{4}\frac{GM}{a} - \frac{GMm}{a} = \frac{1}{2}m\frac{5}{4}\frac{a^2}{r^2}\frac{GM}{a} - \frac{GMm}{r}

\frac{5}{8a}-\frac{1}{a} = \frac{5}{8}\frac{a}{r^2}-\frac{1}{r}

\frac{-3}{8a} = \frac{1}{8r^2}(5a-8r)

3r^2-8ar+5a^2 = 0 which gives r=5a/3 ,a

 

[gneill]

\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}

\frac{1}{2}m\frac{5}{4}\frac{GM}{a} - \frac{GMm}{a} = \frac{1}{2}m\frac{5}{4}\frac{a^2}{r^2}\frac{GM}{a} - \frac{GMm}{r}

The LHS looks fine. But the velocity used on the RHS should be the velocity as obtained via the conservation of angular momentum:

$r \cdot v = a \cdot v_o$ {angular momentum depends on the velocity component perpendicular to the radius vector}

$v = \frac{a}{r} v_o$

$v^2 = \left( \frac{a}{r} \right)^2 v_o^2$ where: $~~~v_o^2 = \frac{GM}{a}$

 

[Tanya Sharma]

Thanks gneill

 

[D H]

Here's the source of your error:

Applying conservation of angular momentum at P and Q

mv_{i}a=mvr

The radial component of velocity does not contribute to angular momentum. This means that firing the gun doesn't change the bullet's orbital angular momentum.