Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: universal gravitation (203-SN1-RE)

Q: Why do astronauts float if there is still gravity?

Because they are in permanent free fall together with their station. At four hundred kilometres of altitude, gravity is still close to ninety percent of its value at the ground: it is precisely what bends their trajectory into a circle. What disappears is the contact force of the floor, not gravity.

Q: Does a higher satellite move faster or slower?

Slower. The orbital speed is the square root of the product of the gravitational constant and the mass of the celestial body, divided by the orbital radius. It therefore decreases as the radius increases. The geostationary satellite, six times farther away than the space station, travels at only three kilometres per second against nearly eight.

Q: Why is gravitational potential energy negative?

Because the reference is chosen at infinity, where it equals zero. Any body closer than infinity therefore has a potential energy below zero, that is, negative. The sign expresses exactly the fact that the body is bound: energy would have to be supplied to take it to infinity.

Q: When can you still use the formula mgh?

Only when the change in altitude is very small compared with the radius of the celestial body, say a few kilometres on Earth. Beyond that, gravity is no longer constant and you must use the full expression, minus G M m over r. For an orbit, the formula is doubly unsuitable since it ignores the kinetic energy.

Q: What is the difference between orbital speed and escape speed?

Orbital speed is the speed that keeps a body on a circle at a given distance. Escape speed is the speed that lets it never come back, by making its mechanical energy zero. At the same distance, the second is exactly the square root of two times the first, that is forty-one percent more.

Q: Does gravity increase if you go down a deep well?

No, it decreases. The layer of matter above you exerts no net force, by the shell theorem: only the mass deeper down counts. Gravity is therefore greatest at the surface, decreases linearly toward the centre where it vanishes, and decreases as one over r squared above.

Gravitation needs only one idea: the gravitational force provides exactly the centripetal acceleration of a circular orbit. From this equality come the orbital speed, the period, Kepler's third law and the mass of celestial bodies. All the rest of the chapter is a matter of energy, with its reference at infinity and its negative sign.

This sheet gathers the errors that cost the most marks on this chapter: the distance measured wrongly, the formula $mgh$ used outside its domain, the sign of the potential energy, and the confusion between the two characteristic speeds of a celestial body. It is written for CEGEP students in the Science program (Sciences de la nature).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Throughout this chapter, $r$ is the distance between the CENTRES, never an altitude. An altitude of $400$ km forgotten next to an Earth radius of $6371$ km hardly changes the drawing and changes everything in the result.

This chapter is part of 203-SN1-RE, Mechanics

The essentials

The formulas and the distance they use

• $F=G\dfrac{m_{1}m_{2}}{r^{2}}$ with $G=6.674\times 10^{-11}$ N·m $^{2}$ /kg $^{2}$ , and $r$ measured between the CENTRES.
• Field of a celestial body at distance $r$ from its centre: $g=\dfrac{GM}{r^{2}}$ . At the surface, $r=R$ ; at altitude $h$ , $r=R+h$ .
• Circular orbit: $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ , so $v=\sqrt{\dfrac{GM}{r}}$ . The mass of the satellite DROPS OUT on both sides.
• Period: $T=\dfrac{2\pi r}{v}$ , hence Kepler's third law $T^{2}=\dfrac{4\pi^{2}}{GM}r^{3}$ , valid for all the satellites of the same celestial body.
• Data to know: $M_{Earth}=5.97\times 10^{24}$ kg, $R_{Earth}=6.37\times 10^{6}$ m, so $GM=3.99\times 10^{14}$ m $^{3}$ /s $^{2}$ .

The force acts on EACH of the two bodies, same magnitude, opposite directions, and

r

is measured between the centres, never between the surfaces.

The product $GM$ is calculated once at the start of a problem and reused everywhere: it is the only way to avoid dragging two powers of ten along on every line.

Several masses: adding vectors

• Each mass attracts the body under study as if it were alone: $F_{i}=G\dfrac{m_{i}m}{r_{i}^{2}}$ , with ITS distance $r_{i}$ , and directed toward it.
• Then you add VECTORS: components $F_{x}$ and $F_{y}$ , then magnitude and direction. Adding the magnitudes is correct only for forces that are aligned and in the same direction.
• The field is superposed in the same way, $\vec{g}=\vec{g}_{1}+\vec{g}_{2}$ : this is what creates the point of zero field between two celestial bodies.

m_{2}

is heavier AND closer:

F_{2}

2.7

times

F_{1}

. The net force, in black, is built head to tail and leans clearly toward

m_{2}

The nearest mass often wins over the heaviest: distance counts squared, mass only to the first power.

The circular orbit: a fall that never ends

• Gravitation is the ONLY force on the satellite: directed toward the centre of the celestial body, it plays the role of the centripetal force, $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ .
• The velocity is tangent to the orbit, hence perpendicular to the force: gravitation bends the trajectory without changing the magnitude of the velocity.
• The acceleration $\dfrac{v^{2}}{r}$ equals the field $g$ at the altitude of the orbit: the satellite and everything inside it fall together, hence the weightlessness felt.

At every point,

\vec{v}

is tangent and

\vec{F}_{g}

aims at the centre of the Earth: the two stay perpendicular, the velocity turns without changing magnitude.

On an orbit diagram, a single force arrow, toward the centre: no “centrifugal force”, no “force of the velocity”.

Energy, with its reference at infinity

• $U=-\dfrac{GMm}{r}$ , reference chosen at infinity. It is NEGATIVE: that is exactly what “bound” means.
• $mgh$ is only an approximation of the change in $U$ for $h\ll R$ . Beyond a few tens of kilometres, it becomes wrong.
• Circular orbit: $K=\dfrac{GMm}{2r}$ , $U=-\dfrac{GMm}{r}$ , so $E=K+U=-\dfrac{GMm}{2r}$ , that is $E=-K$ and $E=\dfrac{U}{2}$ .
• $E<0$ means a closed orbit; $E=0$ means just barely escaping, with zero speed at infinity; $E>0$ means a hyperbolic trajectory.
• Escape speed: $v_{esc}=\sqrt{\dfrac{2GM}{r}}=\sqrt{2}\,v_{orbit}$ , that is $41\%$ more than the orbital speed at the same place.

r=2R

, the energy of the orbit is halfway between

U

and zero:

K

fills half of the well. Doubling

K

, that is multiplying

v

\sqrt{2}

, leads to

E=0

A positive mechanical energy obtained for a satellite in orbit is always a sign error: a bound body necessarily has $E<0$ , and this is the quickest check in the chapter.

The field inside and outside a celestial body

• Shell theorem: a homogeneous spherical shell exerts NO force on a body located inside it.
• Outside, a spherical celestial body acts exactly as if all its mass were concentrated at its centre: $g=\dfrac{GM}{r^{2}}$ , decreasing.
• Inside a homogeneous celestial body, only the mass located LOWER counts, which gives $g=\dfrac{GM}{R^{3}}r$ , increasing linearly.
• The field is therefore MAXIMUM at the surface, zero at the centre, and decreasing above. Going down a well decreases $g$ , it does not increase it.
• Point of zero field between two celestial bodies: it lies where $\dfrac{GM_{1}}{d_{1}^{2}}=\dfrac{GM_{2}}{d_{2}^{2}}$ , so closer to the less massive one.

The field is MAXIMUM at the surface: it rises linearly below, because only the deeper mass acts, and falls back as

1/r^{2}

above. Going down a well therefore reduces the pull of gravity.

The complete curve of $g$ as a function of $r$ regularly comes up as a reasoning question on the final exam: it is remembered by its shape, a straight rise then a descent as $1/r^{2}$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using the altitude as the distance in the law of gravitation

the whole question, 6 marks

What not to write

“The satellite is at $400$ km of altitude, so $g=\dfrac{GM}{(4.00\times 10^{5})^{2}}=2492$ m/s $^{2}$ .”

What to write

“ $r$ is measured from the CENTRE: $r=R+h=6.371\times 10^{6}+4.00\times 10^{5}=6.771\times 10^{6}$ m, so $g=8.69$ m/s $^{2}$ .”

Why: The $1/r^{2}$ law is written for point masses, and a sphere behaves like a point placed at its centre. The altitude is only a difference, never a distance to that point.

2. Believing that the weightlessness of astronauts means the absence of gravity

4 marks on the reasoning questions

What not to write

“In orbit, there is no gravity any more, that is why the astronauts float.”

What to write

“At $400$ km of altitude, $g$ is still $8.69$ m/s $^{2}$ , that is $89\%$ of its value at the ground. The astronauts float because they are in permanent FREE FALL, like their station.”

Why: Without gravity, the station would leave in a straight line. It is precisely gravitation, almost intact, that bends its trajectory into a circle. Weightlessness is an absence of CONTACT force, not of gravity.

3. Using $mgh$ for a change in altitude comparable to the Earth's radius

the whole question, 6 marks

What not to write

“Lifting a $1000$ kg satellite to $400$ km costs $mgh=1000\times 9.8\times 4.00\times 10^{5}=3.92\times 10^{9}$ J.”

What to write

“ $\Delta U=-\dfrac{GMm}{R+h}+\dfrac{GMm}{R}=3.70\times 10^{9}$ J for the altitude alone, and the orbital kinetic energy is still needed. The total is $3.31\times 10^{10}$ J.”

Why: $mgh$ assumes a constant $g$ , which is true only if $h\ll R$ . Here $h$ is already $6\%$ of $R$ , and above all the bulk of the bill is the orbital speed, absent from $mgh$ .

4. Losing the minus sign of the gravitational potential energy

the whole energy question, 5 marks

What not to write

“ $U=\dfrac{GMm}{r}$ , so the mechanical energy of the orbit is $E=\dfrac{3GMm}{2r}$ , a positive value.”

What to write

“ $U=-\dfrac{GMm}{r}$ with the reference at infinity, so $E=\dfrac{GMm}{2r}-\dfrac{GMm}{r}=-\dfrac{GMm}{2r}<0$ .”

Why: The negative sign is not a decorative convention: it encodes the fact that the satellite is BOUND. A positive energy describes a body that escapes, which contradicts the statement.

5. Making the orbital speed depend on the mass of the satellite

4 marks on the reasoning questions

What not to write

“The space station is much heavier than this small satellite, so it must go faster to stay in orbit.”

What to write

“ $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ : the mass $m$ cancels, so $v=\sqrt{\dfrac{GM}{r}}$ depends ONLY on the celestial body and the orbital radius.”

Why: It is the same reason that makes two objects of different masses fall at the same speed: the mass appears on both sides of the equation, in the force and in the inertia.

6. Confusing orbital speed and escape speed

3 marks, and a doubled energy budget

What not to write

“To stay in low orbit, you have to reach $11.2$ km/s.”

What to write

“ $11.2$ km/s is the ESCAPE speed from the ground. A low orbit needs only $7.67$ km/s, about $1.5$ times less. At any given distance, the ratio is exactly $\sqrt{2}$ .”

Why: The two speeds answer two different questions: staying on a circle ( $E<0$ ) or leaving for good ( $E=0$ ). The ratio is exactly $\sqrt{2}$ at a given distance.

7. Believing that a higher satellite moves faster

4 marks on the reasoning questions

What not to write

“The geostationary satellite is much farther away, so it moves much faster than the space station.”

What to write

“ $v=\sqrt{\dfrac{GM}{r}}$ DECREASES as $r$ increases: $7.67$ km/s in low orbit against $3.07$ km/s in geostationary orbit. It takes longer because it goes slower AND covers more distance.”

The curve FALLS: the higher the orbit, the slower the satellite. The geostationary satellite, six times farther away, goes two and a half times slower.

Why: Far from the celestial body, gravity is weaker, so the available centripetal acceleration is weaker, so the speed compatible with a circle of that radius is lower too.

8. Using the radius of the celestial body instead of the radius of the orbit in Kepler's law

the whole question, 5 marks

What not to write

“ $T^{2}=\dfrac{4\pi^{2}}{GM}R^{3}$ with $R$ the radius of the Earth, whatever the altitude of the satellite.”

What to write

“The $r$ in the third law is the radius of the ORBIT, that is $R+h$ , and for an ellipse it is the semi-major axis $a$ .”

Why: The third law relates the period to the size of the trajectory, never to the size of the celestial body. With the radius of the Earth, you would get the period of a satellite grazing the ground, the same for every altitude.

9. Believing that gravity increases as you go below the surface

4 marks on the reasoning questions

What not to write

“The closer you get to the centre of the Earth, the closer you are to its mass, so the greater $g$ is.”

What to write

“The shell located ABOVE you exerts no net force: only the deeper mass counts, so $g=\dfrac{GM}{R^{3}}r$ decreases until it vanishes at the centre.”

Why: The shell theorem is an exact consequence of the $1/r^{2}$ law. The Earth's field is therefore maximum at the surface, a counterintuitive result that comes up on almost every final exam.

Which method to choose

Force or energy

The quantity asked for in the question

If you are asked for an orbital speed, a period, an orbital radius → equate gravitation to the centripetal force, $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$

Example: $v=\sqrt{GM/r}$ , then $T=\dfrac{2\pi r}{v}$
If you are asked for the mass of a celestial body from an observed satellite → Kepler's third law solved for $M$ , that is $M=\dfrac{4\pi^{2}r^{3}}{GT^{2}}$

Example: mass of the Sun from the Earth's orbit, mass of an exoplanet
If you are asked for an energy, a work, an escape speed → balance with $U=-\dfrac{GMm}{r}$ , never $mgh$

Example: energy cost of a change of orbit
If you are asked for a speed at two positions of an ELLIPTICAL orbit → conservation of energy AND of angular momentum, $r_{p}v_{p}=r_{a}v_{a}$

Example: speed at perigee knowing the one at apogee
If you are asked where the field vanishes between two celestial bodies → equate the two fields, $\dfrac{M_{1}}{d_{1}^{2}}=\dfrac{M_{2}}{d_{2}^{2}}$ , with $d_{1}+d_{2}=d$

the point is always closer to the less massive body

A circular orbit links all these quantities: as soon as one is known, the others follow. Writing $GM$ once and for all at the start of your paper avoids recalculating this product at every question.

Which distance to put in the formula

The position of the body relative to the celestial body

If the body is at the surface → $r=R$ , the radius of the celestial body

Example: $g_{0}=\dfrac{GM}{R^{2}}=9.82$ m/s $^{2}$ for the Earth
If the body is at altitude $h$ → $r=R+h$ , and check that $h$ is in metres

Example: $400$ km of altitude gives $r=6.771\times 10^{6}$ m
If the body is inside the celestial body, at depth $d$ → $r=R-d$ , and the field is $\dfrac{GM}{R^{3}}r$ , not $\dfrac{GM}{r^{2}}$

Example: at the centre, $r=0$ and the field is zero
If the orbit is elliptical → $r$ varies, but the semi-major axis $a$ replaces $r$ in Kepler's law and in the energy

Example: $E=-\dfrac{GMm}{2a}$ , valid for any ellipse
If the statement gives the period and asks for the altitude → Kepler to get $r$ , THEN subtract $R$ to get back to the altitude

the geostationary orbit is at $4.22\times 10^{7}$ m from the centre, that is $3.58\times 10^{7}$ m of altitude

Write $r=R+h$ on the first line of every question, even when it seems obvious. This reflex alone eliminates the most frequent trap of the chapter.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating a circular orbit from start to finish

When to use it: The statement gives an altitude or a period and asks for the speed, the period, the energy or the mass of the celestial body.

1 Calculate $GM$ once, and write $r=R+h$ , converting all lengths to metres.
2 Set down the founding equality: gravitation provides the centripetal force, $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ , pointing out that $m$ cancels.
3 Derive $v=\sqrt{\dfrac{GM}{r}}$ , then $T=\dfrac{2\pi r}{v}$ if the period is asked for.
4 For the energy, write the three terms separately: $K=\dfrac{GMm}{2r}$ , $U=-\dfrac{GMm}{r}$ , $E=-\dfrac{GMm}{2r}$ , and check that $E$ is negative.
5 Conclude by giving the results with their units and a commented order of magnitude.

Concluding sentence

“Gravitation provides the centripetal force, so $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ ; the mass of the satellite cancels and $v=\sqrt{\dfrac{GM}{r}}=7.67$ km/s, hence a period $T=\dfrac{2\pi r}{v}=5.54\times 10^{3}$ s, that is $92.4$ minutes.”

The trap: Jumping straight to the formula $v=\sqrt{GM/r}$ without writing the starting equality. The marking scheme almost always rewards the line that identifies gravitation with the centripetal force, because that is the line that contains the physics.

Marking: Typically 1 mark for $r=R+h$ , 2 for the founding equality, 2 for the speed, 2 for the period, 1 for the units and the order of magnitude.

Weighing a celestial body from one of its satellites

When to use it: The statement gives the orbital radius and the period of a satellite, natural or artificial, and asks for the mass of the celestial body.

1 Write the third law in its Newtonian form: $T^{2}=\dfrac{4\pi^{2}}{GM}r^{3}$ .
2 Isolate the mass: $M=\dfrac{4\pi^{2}r^{3}}{GT^{2}}$ , stressing that the mass of the satellite does not appear in it.
3 Convert the period to seconds and the radius to metres BEFORE substituting.
4 Compare the result with a known mass to validate the order of magnitude, for example that of the Earth or the Sun.

Concluding sentence

“The third law gives $M=\dfrac{4\pi^{2}r^{3}}{GT^{2}}$ ; with $r=4.22\times 10^{7}$ m and $T=86{,}164$ s, we obtain $M=5.97\times 10^{24}$ kg, which is indeed the mass of the Earth.”

The trap: Leaving the period in hours or days. The factor $T^{2}$ amplifies the error by squaring it, and a result a trillion times too large sometimes goes unnoticed in a poorly checked scientific notation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign of the mechanical energy

A closed orbit requires $E<0$ . A positive result immediately signals a lost sign on $U$ , before you even check the numbers.

An $E=+2.9\times 10^{10}$ J for a satellite in orbit: the sign of the potential energy has flipped.
The comparison of $g$ with its value at the ground

At $400$ km, $g$ must be about $89\%$ of $9.8$ m/s $^{2}$ . A zero value, or one greater than at the ground, is wrong by construction.

A $g=2492$ m/s $^{2}$ : the altitude was used as the distance instead of $R+h$ .
The order of magnitude of orbital speeds

In low Earth orbit, a speed is of the order of $8$ km/s and a period $90$ minutes. A geostationary satellite travels at $3$ km/s with a period of $24$ h.

A period of $3$ hours for an orbit at $400$ km: too long by a factor of $2$ , the radius was converted wrongly.
The ratio $\sqrt{2}$ between the two speeds

At a given distance, the escape speed is exactly $\sqrt{2}$ times the orbital speed. The ratio can be checked in your head on the two results.

An escape speed of $15$ km/s for an orbital speed of $7.7$ km/s: the ratio is $1.95$ , not $1.41$ .
The mass of the satellite absent from the result

Any expression for orbital speed, period or altitude that still contains $m$ is wrong: this mass always cancels.

A $v=\sqrt{GMm/r}$ : even the dimensions no longer fit, the speed would depend on the mass of the satellite.

The typical problem, taken apart

A low-orbit satellite, from speed to energy cost

A $1000$ kg satellite describes a circular orbit at $400$ km of altitude around the Earth. Take $G=6.674\times 10^{-11}$ N·m $^{2}$ /kg $^{2}$ , $M=5.972\times 10^{24}$ kg and $R=6.371\times 10^{6}$ m.

Determine the orbital speed, the period, the mechanical energy of the orbit, then the energy that had to be supplied to bring it from the ground to this orbit, neglecting the rotation of the Earth and air drag.

The solid line is the surface, the dotted line the orbit: the distance that enters every formula starts from the CENTRE, not from the surface. The drawing exaggerates the gap, which is only

6\%

Step 1

$GM=(6.674\times 10^{-11})(5.972\times 10^{24})=3.986\times 10^{14}$ m $^{3}$ /s $^{2}$ , and $r=R+h=6.371\times 10^{6}+4.00\times 10^{5}=6.771\times 10^{6}$ m.

Why

These two numbers will be used in all the following questions. Calculating them only once avoids dragging two powers of ten along on every line, where they get lost.

Step 2

Gravitation provides the centripetal force: $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ , so $v=\sqrt{\dfrac{GM}{r}}=\sqrt{\dfrac{3.986\times 10^{14}}{6.771\times 10^{6}}}=7.67\times 10^{3}$ m/s.

Why

This is the line that contains all the physics of the chapter, and the one the marking scheme rewards. The mass of the satellite disappears from both sides, which must be said explicitly.

Step 3

$T=\dfrac{2\pi r}{v}=\dfrac{2\pi(6.771\times 10^{6})}{7.67\times 10^{3}}=5.54\times 10^{3}$ s, that is $92.4$ minutes.

Why

Converting to minutes is not cosmetic: it is the only way to recognize a familiar order of magnitude, that of a low orbit, and therefore to validate the calculation.

Step 4

$K=\dfrac{GMm}{2r}=2.943\times 10^{10}$ J, $U=-\dfrac{GMm}{r}=-5.886\times 10^{10}$ J, so $E=K+U=-2.943\times 10^{10}$ J.

Why

The three terms are written separately so that the relation $E=-K=\dfrac{U}{2}$ shows up, because it is the relation that serves as a check and that comes up as a theory question.

Step 5

At the ground, at rest: $E_{ground}=-\dfrac{GMm}{R}=-6.256\times 10^{10}$ J. The energy to supply is therefore $\Delta E=E_{orbit}-E_{ground}=-2.943\times 10^{10}+6.256\times 10^{10}=3.31\times 10^{10}$ J.

Why

You compare two total mechanical energies, never two heights. The difference takes in the climb and the speeding up in one go, which $mgh$ does not do.

Step 6

Comparison with the high school approximation: $mgh=1000\times 9.8\times 4.00\times 10^{5}=3.92\times 10^{9}$ J, that is eight times less than the correct answer.

Why

The factor $8$ shows where the energy of a launch really goes: not into climbing, but into accelerating up to $7.67$ km/s. This is the argument that makes the high school formula unusable here.

Step 7

Checks: $E<0$ as it should be for a closed orbit, $E=-K$ , and $v_{esc}=\sqrt{2}\,v=1.085\times 10^{4}$ m/s stays above the orbital speed.

Why

Three independent checks on structural relations, not on arithmetic. They catch a lost sign or a mislaid factor $2$ , which are the two most frequent errors here.

The conclusion, written out

“The satellite describes its orbit at $7.67$ km/s in $92.4$ minutes, with a mechanical energy of $-2.94\times 10^{10}$ J; bringing it there from the ground cost $3.31\times 10^{10}$ J, most of which goes into reaching speed and not into the climb.”

The classic mistake on this problem: Answering $3.92\times 10^{9}$ J by using $mgh$ . The order of magnitude is off by a factor of $8$ , and above all the answer completely ignores the orbital kinetic energy, which is the real cost of a launch.

Learn by heart

• $F=G\dfrac{m_{1}m_{2}}{r^{2}}$ with $r$ between the CENTRES; at altitude $h$ , $r=R+h$ .
• Circular orbit: $v=\sqrt{\dfrac{GM}{r}}$ , independent of the mass of the satellite, and DECREASING with $r$ .
• Kepler: $T^{2}=\dfrac{4\pi^{2}}{GM}r^{3}$ , with $r$ the radius of the ORBIT, or the semi-major axis for an ellipse.
• $U=-\dfrac{GMm}{r}$ , negative. $mgh$ is valid only for $h\ll R$ .
• Circular orbit: $E=-\dfrac{GMm}{2r}=-K=\dfrac{U}{2}$ , always NEGATIVE.
• $v_{esc}=\sqrt{2}\,v_{orbit}$ at the same distance, that is $11.2$ km/s from the surface of the Earth.
• Shell theorem: $g$ is maximum at the SURFACE, rises linearly below it, falls as $1/r^{2}$ above it.
• $GM_{Earth}=3.99\times 10^{14}$ m $^{3}$ /s $^{2}$ : calculate once, reuse everywhere.

Frequently asked questions

Why do astronauts float if there is still gravity?

Because they are in permanent free fall together with their station. At four hundred kilometres of altitude, gravity is still close to ninety percent of its value at the ground: it is precisely what bends their trajectory into a circle. What disappears is the contact force of the floor, not gravity.

Does a higher satellite move faster or slower?

Slower. The orbital speed is the square root of the product of the gravitational constant and the mass of the celestial body, divided by the orbital radius. It therefore decreases as the radius increases. The geostationary satellite, six times farther away than the space station, travels at only three kilometres per second against nearly eight.

Why is gravitational potential energy negative?

Because the reference is chosen at infinity, where it equals zero. Any body closer than infinity therefore has a potential energy below zero, that is, negative. The sign expresses exactly the fact that the body is bound: energy would have to be supplied to take it to infinity.

When can you still use the formula mgh?

Only when the change in altitude is very small compared with the radius of the celestial body, say a few kilometres on Earth. Beyond that, gravity is no longer constant and you must use the full expression, minus G M m over r. For an orbit, the formula is doubly unsuitable since it ignores the kinetic energy.

What is the difference between orbital speed and escape speed?

Orbital speed is the speed that keeps a body on a circle at a given distance. Escape speed is the speed that lets it never come back, by making its mechanical energy zero. At the same distance, the second is exactly the square root of two times the first, that is forty-one percent more.

Does gravity increase if you go down a deep well?

No, it decreases. The layer of matter above you exerts no net force, by the shell theorem: only the mass deeper down counts. Gravity is therefore greatest at the surface, decreases linearly toward the centre where it vanishes, and decreases as one over r squared above.

Practise it

Corrected exercises: Universal gravitation, 203-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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Revision sheet: universal gravitation (203-SN1-RE)

The essentials

The formulas and the distance they use

Several masses: adding vectors

The circular orbit: a fall that never ends

Energy, with its reference at infinity

The field inside and outside a celestial body

The mistakes that cost marks

1. Using the altitude as the distance in the law of gravitation

2. Believing that the weightlessness of astronauts means the absence of gravity

3. Using mghmghmgh for a change in altitude comparable to the Earth's radius

4. Losing the minus sign of the gravitational potential energy

5. Making the orbital speed depend on the mass of the satellite

6. Confusing orbital speed and escape speed

7. Believing that a higher satellite moves faster

8. Using the radius of the celestial body instead of the radius of the orbit in Kepler's law

9. Believing that gravity increases as you go below the surface

Which method to choose

Force or energy

Which distance to put in the formula

How the answer is expected to be written

Calculating a circular orbit from start to finish

Weighing a celestial body from one of its satellites

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The typical problem, taken apart

A low-orbit satellite, from speed to energy cost

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Frequently asked questions

Why do astronauts float if there is still gravity?

Does a higher satellite move faster or slower?

Why is gravitational potential energy negative?

When can you still use the formula mgh?

What is the difference between orbital speed and escape speed?

Does gravity increase if you go down a deep well?

Corrected exercises: Universal gravitation, 203-SN1-RE

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3. Using $mgh$ for a change in altitude comparable to the Earth's radius