What Initial Speed is Needed for a Cannonball to Escape Earth's Gravity?

In summary, the conversation is about a person struggling with a physics problem and asking for help. They are wondering about launching a cannonball vertically and escaping the Earth's gravitational field. The solution involves using conservation of energy and the force equation to calculate the required initial velocity for the cannonball to escape.
  • #1
Bob__
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0
Hello
I have been struggling with this problem for a whole day, and finally admitting that i can't solve it, so i ask for help :smile:
I am familiar with the escape velocity , and the speed needed to brake into interplanetary space - the Newton cannonball thought experiment. But i was wonder about launching a cannonball perpendicular to the gravitational field - vertical lunch. Ignoring air resistance, and other celestial bodies , what is the initial speed the cannonball must have in order to break trough gravitational influence of the Earth and head to infinity?
I tried , but i got just mathematical understanding of calculus, i never learned it's applications in physics.
I know only that this values change over time like this:
v(t) = Vo-g*t g= M*G/r^2 , and R=ro+V(t) H= Vo*t-(g*t^2)/2 or after brief calculation, H=(Vo^2)/2*g
where M is the mass of earth
r the radius
t-time
ro initial semi major axis of earth
vo- initial velocity
H-max height in homogeneous g-field
because speed tends to reduce for g each second , and g tends to reduce by inverse square of r
there should , in my opinion be a sufficiently large initial velocity with which a verticaly launched projectile would leave the Earth to infinity, as R-> infinity g ->0 , LimV= Vmin and LimH=infinity
How to mathematically express that kind of solution, if there is one?
 
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  • #2
Bob__ said:
I know only that this values change over time like this:
v(t) = Vo-g*t g= M*G/r^2
Your formula v(t) = Vo-g*t is only true when g is constant, so it is wrong for this question.

Probably the easiest way to do the physics is to use conservation of energy. The work required to raise the ball to height h above the center of the Earth is the integral of (force x distance). And force = mass x acceleration, so
work = ##\int_{r_0}^h \frac{mMG}{r^2}dr## where ##r_0## is the Earth's radius.

So the work to reach height h is ##mMG(\frac{1}{r_0} - \frac{1}{h})##.

If the initial kinetic energy ##\frac{mv^2}{2}## is greater than ##\frac{mMG}{r_0}## there is enough energy to reach any height h, so the ball can escape from the Earth's gravitational field.

You can also solve this problem by integrating the "force = mass x acceleration" equation, but this is tricky because the force is a given as a function of the distance, but acceleration is a derivative with respect to time not distance.
 
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  • #3
[itex]\sqrt{2GM/r}[/itex] -and it is the escape velocity
 

Related to What Initial Speed is Needed for a Cannonball to Escape Earth's Gravity?

1. What is a vertical projectile?

A vertical projectile is an object that is launched into the air and follows a curved path due to the force of gravity. It can be any object, such as a ball or a bullet, that is launched with an initial velocity in a vertical direction.

2. What is the equation for vertical projectile motion?

The equation for vertical projectile motion is y = y0 + v0t - 1/2gt^2, where y is the vertical position, y0 is the initial vertical position, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.

3. How is the maximum height of a vertical projectile calculated?

The maximum height of a vertical projectile is calculated using the equation h = v0^2sin^2(theta)/2g, where h is the maximum height, v0 is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity.

4. What is the relationship between the launch angle and the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. The relationship between the launch angle and the range is that the range is maximum when the launch angle is 45 degrees. This means that the projectile is launched at an angle of 45 degrees from the ground.

5. How does air resistance affect the trajectory of a vertical projectile?

Air resistance can have a significant effect on the trajectory of a vertical projectile. It can cause the projectile to slow down and deviate from its expected path, leading to a shorter range and lower maximum height. The amount of air resistance depends on the shape and size of the projectile, as well as the density of the air and the projectile's velocity.

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