Physics Bootcamp

(a) We use the formulas given.

(b) We use the formulas given.

(a) I have stated above that a spherical object “acts” same as a point particle as far as gravitation outside the sphere is concerned. Therefore, the force between the spherical lead balls can be easily computed giving the following magnitude.

(b) Since people are not spherical objects, and since their separation is not too great compared to their sizes, we cannot use the same form of the law of gravitation as given for the point particles. If we use the formula given, we will only get some rough estimate, not an exact answer for the magnitude of the force.

(c) In this part, the people are separated by more than 10-times their sizes. Therefore approximating them as spheres will not give us too different a result than an exact result. Although the result of spherical approximation for a person will not give an exact result, but we expect the result to be more precise here than it was in part (b). The magnitude of the force is

(d) Although, the satellite is not a spherical object, it is far away from the Earth compared to the size of the satellite. Therefore, we can treat the satellite as a point mass. This gives the following for the magnitude of the force on the satellite.

(a) The magnitude of the gravitational force is

Note that I separated out the powers of 10. This often helps prevent numerical error.

(b) Dividing by mass of Earth will give the magnitude of acceleration.

(c) Dividing by mass of Sun will give the magnitude of acceleration.

A geocentric orbit means that the satellite has the same time period as the time period of the rotation of Earth. When a satellite is in a geocentric orbit, the satellite appears at the same spot above the Earth. These satellites are used for telecommunications.

Let \(h \) be the altitude above Earth, and \(R_E\) the radius of the Earth. Then, the radius of the circular motion of the satellite is \(R_E+h\text{.}\) Since the satellite covers a distance of one circumference \(2\pi(R_E+h)\) in time \(T = 1\text{ day}\text{,}\) the average speed of the satellite must be

We will assume that the speed of the satellite is constant. This means that the satellite would be in a uniform circular motion and the magnitude of the centripetal acceleration of the satellite would be

The centripetal acceleration is the net acceleration of the satellite in this case. The net force on the satellite is the force from the Earth. Therefore, we can write the following using the magnitudes of the force on the satellite and the acceleration of the satellite.

We can solve this equation for the altitude of the satellite in a geocentric orbit with the following result.

(a) We first find the forces on 1 by 2 and 3 separately and then add them vectorially to obtain the net force on 1. Lets denote the magnitudes of these forces as \(F_{21}\) and \(F_{31}\text{.}\) The magnitudes of these forces are

Looks like we should use the unit of \(10^{-7}\text{ N} \text{.}\) We will call this unit \(\text{N*}\) to save space.

To add these forces, we will use analytic method. For that we inroduce axes. \(F_{21}\) is towards positive \(x\)axis and \(F_{31}\) towards positive \(y\)axis. From the figure we get

\begin{align*} F_\text{net} \amp = \sqrt{ F_{21}^2 + F_{31}^2 }\\ \amp = \sqrt{3.3^2 + 10.0^2} = 10.5 \text{ N*}\\ \amp = 10.5\times 10^{-7}\text{ N}. \end{align*}

We use angle \(\theta\) to indicate the direction. The counterclockwise angle from the positive \(x\) axis is

Lets denote the magnitudes of forces on 2 by \(F_{12}\) and \(F_{32} \text{.}\) We have already worked the magnitude of the force between 1 and 2, which we had called \(F_{21}\text{.}\)

The other force on 2 is

That is \(F_{32} = 4.0\text{ N*}\text{.}\) To add these two forces we use the components.

Therefore, the components of the net force are

The magnitude of this force is

\begin{equation*} F_{\text{net}} = \sqrt{ 6.9^2 + 1.8^2 } = 7.1\text{ N*} = 7.1\times 10^{-7}\text{ N}. \end{equation*}

The direction is in the second quadrant with clockwise angle with respect to the negative \(x\) axis give by

\begin{equation*} \theta = \tan^{-1}\left(\dfrac{1.8}{6.9} \right) = 14.6^\circ. \end{equation*}

(c) We have worked out the magnitudes \(F_{13}\) and \(F_{23}\) already as \(F_{31}\) and \(F_{32}\text{.}\)

We also know their components as per Newton's third law.

Therefore, the net force on 3 has the following components.

The magnitude of this force is

\begin{equation*} F_{\text{net}} = \sqrt{ 3.^2 + 11.8^2 } = 12.2\text{ N*} = 12.2\times 10^{-7}\text{ N}. \end{equation*}

The direction is in the fourth quadrant with clockwise angle with respect to the positive \(x\) axis give by

\begin{equation*} \theta = \tan^{-1}\left(\dfrac{11.8}{3.6} \right) = 73^\circ. \end{equation*}

We place the sphere with its center at the origin and the rod along \(x\) axis as shown in Figure 12.2.9. Now consider the force on the sphere by an infinitesimal element of the rod between \(x\) and \(x+dx\text{.}\) The mass in the element \(dx=\lambda dx\text{,}\) where \(\lambda = M/L\text{,}\) the linear mass density.

We need to integrate this from \(x=R\) to \(x=R+L\text{.}\)

We place the mass of the sphere at its center and examine the force on this mass by two infintesimal elements of rod symmetrically placed as shown in Figure 12.2.12. Note that the direction of the two forces is such that the sum of their horizontal components is zero. Therefore, we need to find only the vertical component.

The \(y\) component of force on the sphere by a small element between \(x\) and \(x + dx \) is

The infinitesimal mass, \(dM = (M/L)dx \) .We integrate this from \(x = -L/2\) to \(x = L/2\text{.}\)

Therefore, the force on the sphere has magnitude \(\dfrac{ G_N m M }{ D \sqrt{D^2 + L^2/4} }\) and direction towards the center of the rod.

Looking at force from two elements of the ring that are at the diagonally opposite sides shows us that only the vertical component is non-zero. The magnitude of the force from an element of the ring on the sphere is

\begin{equation*} dF = G_N \dfrac{m\, (M/L) dl}{ D^2 + R^2} \end{equation*}

Let \(z\) axis be pointed up. Then, \(z\) component of the gravitational force from an element of the ring of size \(dl\) is

which is independent of the location of the element on the ring. When we integrate this aroiund the ring, we will get

Therefore, the magnitude of the force on the sphere is \(\dfrac{G_N m M D}{ \left( D^2 + R^2 \right)^{3/2} } \) towards the center of the ring.

We will build the solution of the disk from the solution of a ring between \(r\) and \(r+dr\text{.}\) The mass in the infinitesimal ring will be

\begin{equation*} dM = \dfrac{M}{\pi R^2}\, 2\pi r dr, \end{equation*}

since area of the ring is \(2\pi r dr\text{.}\) From symmetry, we also know that only the vertical component of the force is non-zero. Let \(dF_z\) denote the \(z\) component of the ring. From the ring solution we have

Now, we integrate this from \(r=0\) to \(r=R\) to get the contribution of the entire disk.

The magnitude is the above quantity without the sign and direction is towards the center of the ring.