Radius of a circle in relation to banking angle

A question was posed in Physics as follows:

An airplane is flying in a horizontal circle at a speed of 480 km/hr. If its wings are titled at a 40° angle relative to the horizontal, what is the radius of the circle (in km) in which the plane is flying? Assume that the required force is proved by the lift which acts perpendicular to the wing surface.

Now let’s identify the information that you need to know:

Velocity = 480 km/hr (133.3 m/s) (you need to convert this for the calculations to work right)
Angle = 40 degrees
Gravitational Acceleration = 9.8 m/s^2

What is important to note in this problem is that the force is modified  by the 40 degree angle meaning that the circle can be tighter because the force is higher.  The easiest way to do this is to use a derived formula.  I’m going to shorten it a bit for time sake but you should be aware that it comes from this combination of horizontal and vertical forces:

Horizontal - F_n\sin{\theta} = m\frac{v^2}{r}
Vertical - F_n\cos{\theta} - mg = 0

This relationship gives us the following formula:

\tan{theta} = \frac{v^2}{rg} \equiv r = \frac{v^2}{g \tan{theta}}

Now we apply this to the known values to get our turning radius:

r = \frac{133.3 m/s}{9.8 m/s^2 \tan{40}} \equiv r = 2160.83 m \equiv r = 2.16 km

As always, please feel free to comment and or correct any conceptual or mathematical errors on my part.  Thanks!

Caclulating distance related to tangential velocity

A question was posed in Physics as follows:

A 60 kg ball is tied to the end of a 50 cm long mass-less string and swung in a vertical circle. The center of the circle is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack. If the string is released at the instant the ball is at the top of the loop, how far to the right (in meters) does the ball hit the ground?

Okay, now we need to look up the key items in this question:

mass = 60 kg
radius = .5 m (50 cm)
height = 1.5 m (150 cm)
gravitational acceleration = 9.8 m/s^2
vertical circle (tossed overhead with release at the peak)
minimum speed for no slack ( t must be greater than or equal to 0)

Now, what this tells us is that we need to a) find the tangential velocity at the top of the circle, b) find the minimum tension on the rope that it won’t go slack at the top, c) the time that it takes the ball to fall back to the ground from it’s highest point (this will allow us to calculate distance on the x-axis).  It is important to note that for this equation mass is extra information.

Let’s solve for the time from the top of the circle to the ground since that’s the easiest to get right away (a quick note, the peak is the height to the middle plus the radius of the circle, so, 2 meters):

T = \sqrt{\frac{2d}{a}} \equiv T = \sqrt{\frac{2(2 m)}{9.8 m/s^2}} \equiv T = 0.64 s

Conceptually we need to realize that the velocity of the object at the top of the circle will be related to the gravity, radius and tension.  Since we want the minimum speed possible, that means the tension must be 0 or greater but not less than 0.  That means that the tension goes out of the problem because we’re just dealing with gravity and the radius of the circle since tension is 0.  That means we can figure out velocity based on the relationship between gravity and the radius of the circle:

v = \sqrt{rg} \equiv v = \sqrt{2 m*9.8 m/s^2} \equiv v = 2.21 m/s

Now that we have the velocity and the time that it takes for the object to get to the ground we use the very simple formula d = vt in order to find the distance the object travels on the x-axis.

d = vt \equiv d = (2.21 m/s)(0.64 s) \equiv d = 1.41 m

Ta-da!  Hopefully this is somewhat helpful in explaining how to get this answer.  Please feel free to comment and or correct me if you find errors.

How to Calculate Weight on Another Planet

*This post re-published from cache – August 28th 2009*

In working on answering a question today, I was rather perturbed by the fact that I was not able to find a definitive answer to a question that I felt should be well defined.  That question was: If I weigh 100lbs on Earth, how much would I weigh on Saturn?

Simple question right?  Not by what I found online.  There are a number of different sites that can be used to get a quick ( and possibly incorrect) answer.  I found the sites that I visited in search of an answer to be at the least, inaccurate and at the most, dead wrong.  The answers ranged from around 55lbs to about 115lbs.  On top of this, the method for calculating weight on another planet was based on the gravitational acceleration at the surface.  This is something that is defined already depending on the site or information you look at, but again, the values varied significantly.

Here’s an easy way to do it and get a reliable answer:

Find the gravitational acceleration, or more specifically the ratio of acceleration in relation to Earth, where Earth is a value of 1.  Now, find a table online with the planet’s ratio that you want to calculate.  For Saturn I used 1.065 (1.07 significant digits) found here.  Simply cross-multiply to get your answer:

\frac{100 lbs.}{1} = \frac{x}{1.07}

100 * 1.07 = x * 1

x = 107 lbs

Now that we have the easy answer I think it is important to understand that without already having the gravitational ratio this is still calculable by obtaining the gravitational ratio on our own, as shown below.

Before we begin, it should be noted that the radius of a planet type body varies from point-to-point.  This means that your answer will also vary depending on where you measure from.  This could account for some of the discrepancies that I found on other sites.  For example, you will be heavier at either of the poles and lighter at the equator.  This is due to the bulge at the equator caused by rotation.  This increases distance from the center of mass and reduces the gravitational forces on you slightly, causing you to weigh less.  For our purposes, we will use the mean radius vs. a specific measurement.

First, we need my mass in kilograms (kg), rather than weight in pounds (lbs) so we use the formula here:  \frac{x}{2.2} = x kg

\frac{100}{2.2} = 45.5 kg

Great, now that we have the mass, we can move forward.  We will need additional information at this point.  The formula that will allow us to make further calculations, Newton’s Law of Universal Gravitation.


It should be noted that the value for r should be in meters, the values of m_1 and m_2 in kilograms and the resultant value of F will be in Newtons.

This formula will allow us to calculate the force exerted between objects of a given mass, at a given distance from the center of gravity.  In turn, allowing us to calculate the respective weight.  For this formula to work, we will need additional statistics.  Since we are calculating for Saturn specifically, I took the values found from the NASA stats page linked earlier.

  • G = The Gravitational Constant (6.67*10^-11 N^2x m^2/kg^2)
  • m_1 = Object 1 Mass, Saturn (5.68*10^26 kg)
  • m_2 = Object 2 Mass, Person (45.5 kg)
  • r = Mean Radius, Saturn (5.82*10^7)

Now that we have all the variables needed, we can use Newton’s formula to get the result:

F_1 = (6.67*10^-11)\frac{(5.68*10^26)(45.5)}{(5.82*10^7)^2}

F_1 = \frac{1.72*10^18}{3.39*10^15}

F_1= 508N

Now we have the gravitational force (F) for Saturn.  In order to make a comparison based on weight, we will also need to do the same calculations for Earth.  Since we already have the gravitational constant and our calculated mass, we need to acquire Earth’s mean radius and it’s mass in kilograms.  These can both be found through the site linked earlier.

  • G = The Gravitational Constant (6.67*10^-11 N^2x m^2/kg^2)
  • m_1 = Object 1 Mass, Earth (5.97*10^24 kg)
  • m_2 = Object 2 Mass, Person (45.5 kg)
  • r = Mean Radius, Earth (6.37*10^6)

Once again we can use Newton’s formula to calculate the value of F:

F_2 = (6.67*10^-11)\frac{(5.97*10^24)(45.5)}{(6.37*10^6)^2}

F_2 = \frac{1.81*10^16}{4.06*10^13}

F_2= 446N

Now that we have both values we can compute the ratio:

R = \frac{F_1}{F_2}where F_1 is the value of Saturn and F_2 is Earth.

R = \frac{508}{446}

R = 1.14

Curiously enough, the ratio arrived at in this equation is higher than the 1.065 cited in the NASA document.  I am assuming this is due to differences in the value used for each planet’s radius.  I will see what I can do about further investigating this discrepancy and amend my post if needed.

Now, to finish our process.  We once again cross-multiply using 100lbs to 1 for the value on Earth.

\frac{100 lbs.}{1} = \frac{x}{1.14}

100 * 1.14 = x * 1

x = 114 lbs

In conclusion, the variations in answer can be explained by taking into account the differences in radius values used in the calculations.  Polar radius, Equatorial radius and mean radius will all give values slightly different from one another in a difference of P_r < m_r < E_r.

With our calculations it can be assumed that we have a possibility for error.  Based on this information and given a discrepancy of a significant 7.4 lbs, this could account for roughly 7-13% variance in answers to this question.  That said, make certain to note which radius measurement is being asked for when calculating these types of problems.  It could mean the difference between a right and wrong answer.