Newton's Law of Gravitation Feynman Lectures

[MarkFarrell82]

Hi all,
I've been reading the Feynman Lectures on Physics and I've stumbled on something. I understand the theory but not how they arrived at the answer. It's to do with firing a bullet from a gun and working out the speed it would need to travel in a curve around the Earth's surface in order to be at the same height that it started out. They prove it using plane geometry which has confused me because as far as I understand it the bullet would need to travel in an arc for which ultimately the equation will need an angle. The answer is about 5 miles a second. Please can someone explain to me how they would approach this problem assuming that the raduis of the Earth is 4000miles and the an object will fall 16ft/sec under the influence of gravity?

Thanks
Mark

 

[Gnosis]

If you accelerate a projectile to the appropriate tangential velocity, it will possesses insufficient velocity to escape the gravitational field just as it will be a velocity too great to allow the projectile’s descent to the planet’s surface. The projectile’s tangential velocity per second is just sufficient to counter the projectile’s rate of vertical free-fall descent per second per the rate of gravitational attraction at the given radius thereby allowing the projectile to maintain an indefinite circular orbit (assuming no frictional losses).

The following yields the velocity (v) required to orbit per a given radius (r) and a given heavenly body’s Mass (M), with ‘G’ being the Gravitational constant, 6.67e-11:

v = sqrt(GM / r)

sqrt(6.67e-11 * 5.976e+24 kg) / 6,377,569.11 meters) = 7,905.7 m/s

Or, you could apply Newton’s Centripetal Force. The earth’s gravitational acceleration (a) at its surface is 9.8 m/s^2. Applying the same Earth radius (r) of 6,377,569.11 meters, we derive a required tangential velocity (v) of:

v^2 = ar

Hence, the required tangential orbital velocity (v) is,

v = sqrt(ar)

sqrt(9.8 m/s^2 * 6,377,569.11 meters) = 7,905.7 m/s

I applied precision numerical values to demonstrate the interchangeable precision of the two equations. I hope you found this helpful.

 

[MarkFarrell82]

Hi Gnosis
Thanks for the response. Only just had the chance to look at it.
Mark

 

[Gnosis]

Hi Gnosis
Thanks for the response. Only just had the chance to look at it.
Mark

It was my pleasure, Mark.

 

[PJC01]

I can provide some insight into this problem based on Newton's Law of Gravitation. This law states that any two objects with mass will exert a gravitational force on each other, and the strength of this force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In simpler terms, the closer two objects are and the more massive they are, the stronger the gravitational force between them will be.

In the case of the bullet fired from a gun, the Earth's gravitational force is acting on it as it travels through the air. This force causes the bullet to fall towards the Earth at a rate of 16 feet per second squared. However, the bullet also has a forward velocity from being fired from the gun. In order for the bullet to maintain a constant height and travel in a circular path around the Earth, its forward velocity must be great enough to counteract the force of gravity pulling it towards the Earth.

To calculate this velocity, we can use the equation for centripetal force, which is the force required to keep an object moving in a circular path. This equation is F = mv^2/r, where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

In this case, the centripetal force is equal to the force of gravity pulling the bullet towards the Earth. So we can set F = mg, where g is the acceleration due to gravity (16 feet per second squared). We also know the radius of the Earth (4000 miles) and we can assume the mass of the bullet is negligible compared to the Earth's mass.

Putting all of this together, we can solve for the velocity (v) using the equation mg = mv^2/r. Rearranging, we get v = √(rg), where r is the radius of the Earth and g is the acceleration due to gravity.

Plugging in the given values, we get v = √(4000 miles * 16 feet per second squared) ≈ 5 miles per second. This means that for the bullet to travel in a circular path around the Earth at a constant height, it would need to have a velocity of approximately 5 miles per second.

I hope this helps to explain the approach to this problem and how Newton's Law of Gravitation can be applied to solve it. Keep