laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

circular orbits

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

G = 6.67 \times 10^{11} \left[N \frac{m^2}{kg^2}\right] = 6.67 \times 10^{11} \frac{m^3}{kg s^2}

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

G = 6.67 \times 10^{11} \left[N \frac{m^2}{kg^2}\right] = 6.67 \times 10^{11} \frac{m^3}{kg s^2}

F = ma

F = \frac{mMG}{R^2} => \frac{MG}{R^2} = a

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

G = 6.67 \times 10^{11} \left[N \frac{m^2}{kg^2}\right] = 6.67 \times 10^{11} \frac{m^3}{kg s^2}

F = ma

F = \frac{mMG}{R^2} => \frac{MG}{R^2} = a

how about F = mg?

laws of gravity

back to uniform circular motion

F = \frac{mMG}{R^2}

\vec{F} = \frac{mMG}{R^2}\hat{r}

circular orbits

G = 6.67 \times 10^{11} \left[N \frac{m^2}{kg^2}\right] = 6.67 \times 10^{11} \frac{m^3}{kg s^2}

F = ma

F = \frac{mMG}{R^2} => \frac{MG}{R^2} = a

Gravitational Potential Energy

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

\infty

F =\int_R^\infty \frac{mMG}{R^2} = -\frac{mMG}{R}|_R^\infty = \frac{mMG}{R} = W

\Delta U = -W

U(R) -U_\infty = U(R) - 0 = -\frac{mMG}{R}

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

Gravitational Potential Energy

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

\infty

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

U(R) = -\frac{mMG}{R}

escape velocity

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

U(R) = -\frac{mMG}{R}

v_{esc}

ME = K + U = \frac{1}{2}mv^2 - \frac{mMG}{R}

ME_\infty = K + U = 0

ME = ME_\infty = 0

v_{esc} =\sqrt{\frac{2MG}{R}}

escape velocity

circular orbits

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{circ}

if the orbit is circular

a = \frac{v^2}{R}

a = \frac{F}{m} = \frac{GM}{R^2}

\frac{v^2}{R} = \frac{GM}{R^2}

v_{circ} = \sqrt{\frac{GM}{R}}

escape velocity

circular orbits

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{esc}

v_{circ} < v < v_{esc} =>\mathrm{elliptical~orbit}

escape velocity

circular orbits

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{esc}

v_{esc} =\sqrt{\frac{2MG}{R}}

escape velocity

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

U(R) = -\frac{mMG}{R}

v_{esc}

v_{esc} =\sqrt{\frac{2MG}{R}}

escape velocity

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

U(R) = -\frac{mMG}{R}

v_{esc}

only depends on the

mass and radius of the earth!

v_{esc} =\sqrt{\frac{2MG}{R}}

v_{escE} = 11,190~m/s

escape velocity

F = \frac{mMG}{R^2}

circular orbits

F = \frac{mMG}{R^2}

R

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

U(R) = -\frac{mMG}{R}

v_{esc}

only depends on the

mass and radius of the earth!

v_{esc} =\sqrt{\frac{2MG}{R}}

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

U = - \frac{mMG}{2a}

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

dA = \frac{1}{2}r r\cdot d\theta

the wedges are triangles after all

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

dA = \frac{1}{2}r r\cdot d\theta

the wedges are triangles after all

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt}

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

dA = \frac{1}{2}r r\cdot d\theta

the wedges are triangles after all

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt}

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt} = \frac{1}{2}r^2 \omega

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

dA = \frac{1}{2}r r\cdot d\theta

the wedges are triangles after all

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt}

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt} = \frac{1}{2}r^2 \omega

L = r p \sin{\theta} = r mv = m r^2 \omega

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

1 - the orbit (of planets) are ellipses

2 - equal areas of the ellipse (wedges) are spanned in equal time

3 -

T^2 \propto a^3

dA = \frac{1}{2}r r\cdot d\theta

the wedges are triangles after all

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt}

\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt} = \frac{1}{2}r^2 \omega

L = r p \sin{\theta} = r mv = m r^2 \omega

\frac{dA}{dt} = \frac{L}{2m}

dA/dt constant means conservation of angular momentum

elliptical orbits

how far is Neptune?

d = 4,568,488,662 kilometers

c = 299,792 km/s

how much longer is the period of its orbit?

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

d[sec] = d / v \\ dt = 4.6 \times 10^12 [m] / 3\times 10^{8} [m/s] = \\ = 1.5 \times 10^4 s = 4 hours

elliptical orbits

how far is Neptune?

d = 4,568,488,662 kilometers

c = 299,792 km/s

how much longer is the period of its orbit?

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

d[sec] = d / v \\ dt = 4.6 \times 10^12 [m] / 3\times 10^{8} [m/s] = \\ = 1.5 \times 10^4 s = 4.1~ hours

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

v = v_{esc}

parabolic

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

Because the velocity does not depend on the mass of the orbiting star, but only on the mass the body around which the star orbits, we can detect invisible masses by studying orbits!

The mass of the Black Hole at the center of our own Galaxy was measured

elliptical orbits

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

laws of gravity

Because the velocity does not depend on the mass of the orbiting star, but only on the mass the body around which the star orbits, we can detect invisible masses by studying orbits!

The mass of the Black Hole at the center of our own Galaxy was measured

Andrea Ghez is the fourth woman to win a Nobel Prize in physics (2020).

Donna Strickland won for the discovery that led to short-pulse high-intensity lasers (2018)

Maria Goeppert Mayer discoveries concerning nuclear shell structure (1963)

Marie Curie studies of the spontaneous radiation (1903)

Blach hole physics

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{esc} =\sqrt{\frac{2MG}{R}}

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{esc} =\sqrt{\frac{2MG}{R}}

schwarzschild radius:

v_{esc_{BH}} =\sqrt{\frac{2M_{BH}G}{R_S}} = c

c is the speed of light, and nothing can go faster!!

Blach hole physics

so nothing can escape!

M_{earth} = 5.9 \times 10^{24} kg\\ R_{earth} = 6.378 \times 10^6 m\\ M_{sun} = 2 \times 10^{30} kg\\ R_{earth-sun} = 1.5 \times 10^{11} m

v_{esc} =\sqrt{\frac{2MG}{R}}

schwarzschild radius:

v_{esc_{BH}} =\sqrt{\frac{2M_{BH}G}{R_S}} = c

c is the speed of light, and nothing can go faster!!

Blach hole physics

so nothing can escape!

really this is a "dark star"

to describe a black hole we need general relativity

special relativity

The laws of physics are the same for all observers in any inertial frame of reference relative to one another (principle of relativity).
The speed of light in a vacuum is the same for all observers, regardless of their relative motion or of the motion of the light source.

Relativity of simultaneity: Two events, simultaneous for one observer, may not be simultaneous for another observer if the observers are in relative motion.
Time dilation: Moving clocks are measured to tick more slowly than an observer's "stationary" clock.
Length contraction: Objects are measured to be shortened in the direction that they are moving with respect to the observer.
Maximum speed is finite: No physical object, message or field line can travel faster than the speed of light in a vacuum.
Mass–energy equivalence: E = mc^2, energy and mass are equivalent