*Notes to a video lecture on http://www.unizor.com*

__Problems on Gravity__

*Problem 1*

Gravitational potential of a spherical gravitational field around a point-mass

*at a distance*

**M***from it is defined as the work performed by gravity to bring a probe object of a unit mass from infinity to this point and is expressed as*

**r**

**V**_{r}= −G·M /rWhy is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass

*, but only on a distance itself from the source of gravity?*

**M***Solution*

Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.

In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work

*d*performed by the force of gravity

**W***during the movement of a probe object, described by the infinitesimal displacement*

**F***d*, is a scalar product of these two vectors:

**S***d*

**W = F·**d**S**Note that the vector of gravitational force

*is always directed towards the source of gravity.*

**F**Since a displacement vector

*d*can be represented as a sum of radial (towards the source of gravity)

**S***d*and tangential (perpendicular to radius)

**S**_{r}*d*components, the above expression for a differential of work can be written as

**S**_{t}*d*

**W = F·****(**

*d*

**S****d**_{r}+**S**_{t}**)**

*=*

= F·= F·

*d*

**S****d**_{r}+ F·**S**_{t}The second component in the above expression is a scalar product of two perpendicular vectors and is equal to zero. That's why we can completely ignore tangential movements, when calculating the work done by a central gravitational field, as not contributing to the amount of work. The total amount of work will be the same as if our probe object moved along a straight line towards the source of gravity and stopped at a distance

*from it.*

**r***Problem 2*

Given two point-masses of mass

*each, fixed at a distance*

**M***from each other.*

**2R**Prove that the gravitational potential of a gravitational field produced by both of them at each point on a perpendicular bisector between them equals to a sum of individual gravitational potentials of these point-masses at this point, as if they were the only source of gravitation. In other words, prove that gravitational potential is additive in this case.

*Solution*

Let's draw a diagram of this problem (you can download it to display in a bigger format).

Our two point-masses are at points

*A*and

*B*, the probe object is at point

*D*on a perpendicular bisector of a segment

*AB*going through point

*C*.

The force of gravity towards point

*A*is a segment

*DE*, the force of gravity towards point

*B*is a segment

*DF*.

We will calculate the potential of a combined gravitational field of two point-masses at point

*D*, where the probe object is located.

Let's assume that the segment

*CD*equals to

*.*

**h**The magnitude of each gravitational force equals to

**F = G·M·m /(h**^{2}+r^{2})Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector

*CD*, another (red) going horizontally parallel to

*AB*.

Vertical components of these two forces will add to each other, as equal in magnitude and similarly directed downwards on a drawing, while horizontal ones will cancel each other, as equal in magnitude and opposite in direction to each other. So, the combined force acting on a probe object is a sum of vertical components of gravitational forces with a magnitude

**F**_{tot}= 2·G·M·m·sin(φ)/(h^{2}+r^{2})Since

*,*

**sin(φ) = CD/AD***[(*

**sin(φ) = h /***)*

**h**^{2}+r^{2}*]*

^{1/2}

**F**_{tot}= 2·G·M·m·h /(h^{2}+r^{2})^{3/2}If the gravitational field pulls a probe object along the perpendicular bisector of a segment

*AB*from infinity to a distance

*from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment*

**h***is changing, according to a similar formula:*

**x**

**F**_{tot}(x) = 2·G·M·m·x /(x^{2}+r^{2})^{3/2}To calculate work performed by a gravitational field pulling a probe object from infinity to height

*above the segment*

**h***AB*, we have to integrate

**W**d_{tot}= ∫_{[∞;h]}F_{tot}(x)·**x**It's supposed to be negative, since the direction of a force is opposite to a positive direction of the coordinate axis, we will take it into account later.

**W**d_{tot}= ∫2GMm·x·**x /(x**^{2}+r^{2})^{3/2}(within the same limits of integration [

*])*

**∞;h**This integral can be easily calculated by substituting

*,*

**y=x**^{2}+r^{2}*,*

**2·x·**d**x =**d**y**infinite limit of integration remaining infinite and the

*limit transforming into*

**x=h***. Now the work expression is*

**y=h**^{2}+r^{2}

**W**d_{tot}= ∫G·M·m·y^{−3/2}·**y**with limits from

*to*

**y=∞***.*

**y=h**^{2}+r^{2}The indefinite integral (anti-derivative) of

*is*

**y**^{−3/2}*.*

**−2·y**^{−1/2}Therefore, the value of integral and the work are

**W**_{tot}= −2·G·M·m·(h^{2}+r^{2})^{−1/2}For a unit mass

*this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance*

**m=1***from a midpoint between them along a perpendicular bisector*

**h**

**V**_{tot}= −2·G·M·(h^{2}+r^{2})^{−1/2}At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to

**V**_{single}= −G·M·(h^{2}+r^{2})^{−1/2}As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.

**IMPORTANT NOTE**

With more cumbersome calculations this principle can be proven for any two (not necessarily equal) point-masses at any point in space (not necessarily along the perpendicular bisector). This principle means that

**gravitational potential is additive**, that is the

__gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials__.

*Problem 3*

Express mass

*of a spherical planet in terms of its radius*

**M***and a free fall acceleration*

**R***on its surface.*

**g***Solution*

Let

*be a mass of a probe object lying on a planet's surface.*

**m**According to the Newton's 2nd Law, its weight is

**P = m·g**According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is

**F**_{gravity}= G·M·m /R^{2}Since the force of gravitation is the weight

*,*

**F**_{gravity}= P

**m·g = G·M·m /R**^{2}from which

**M = g·R**^{2}/G*Problem 4*

Express gravitational potential

*of a spherical planet on its surface in terms of its radius*

**V**_{R}*and a free fall acceleration*

**R***on its surface.*

**g***Solution*

From the definition of a gravitational potential on a distance

*from a source of gravity*

**R**

**V**_{R}= −G·M /RUsing the expression of the planet's mass in terms of its radius

*and a free fall acceleration*

**R***on its surface (see above),*

**g**

**M = g·R**^{2}/GSubstituting this mass into a formula for potential,

**V**_{R}= −G·g·R^{2}/(G·R) = −g·R
## No comments:

Post a Comment