[chris (OP)]

How did scientists measure the mass of the Earth and other planets?

How did Newton know he'd got his maths right?

[alancalverd]

Using F = GmM/r² you can calculate the force on a falling object of mass m in terms of M, the mass of the earth, and G, which we assume to be a universal constant.

As F = ma, we can measure the acceleration of a falling object or the period of a pendulum to get a value for F/m

Cavendish's experiment is a classic procedure for measuring G.

Hence M.

The Inverse Square Law is "obvious" if you think of gravity like llight, propagating in straight lines, so it's a good starting guess. Using the ISL, you can show that the orbit of a planet around a star is necessarily elliptical. Observation shows this to be correct, and you can preedict the orbit of a comet to a very pleasing degree of accuracy . Deviations from the idealised ellipse will be caused by the presence of other massive bodies, so we gradually accumulate "corrections" to our astronomical observations that give us estimates of the mass of these other objects.

Like Einstein and indeed every other scientist worthy of the name, Newton made some predictions which were tested by observation, and thus he (or his successors) demonstrated the validity of the mathematics. This is what distinguishes science from computer modelling, economics, philosophy, politics, or religion.

[jarvisss]

I think that more interesting how they measure distance in the space between planets.
Don't want to quote big pieces of the text
here is the video youtube.com/watch?v=_Lsj-Hz-NS4

[Halc]

How did scientists measure the mass of the Earth and other planets?

How did Newton know he'd got his maths right?

This is actually a very good question.  It is sort of a chicken/egg problem, which requires an answer to know the answer.  To illustrate:

Using F = GmM/r² you can calculate the force on a falling object of mass m in terms of M, the mass of the earth, and G, which we assume to be a universal constant.

That doesn’t work.  We’re trying to compute at least a rough G and M here.  We don’t know either of them yet.  We do know force F is 9.8 newtons for a 1KG mass.  We can assume we know r.  We therefore know the product of G and M, but not either separately.

As F = ma, we can measure the acceleration of a falling object or the period of a pendulum to get a value for F/m

F=ma works (F 9.8 = 1 (mass) * 9.8 m/sec acceleration), but that doesn’t yield either mass of Earth M nor G, which are the two things we’re trying to determine here.
The pendulum thing is a function of acceleration (9., not of the mass of Earth.  Put a pendulum in a rocket accelerating at that rate and it will have the same period as here on Earth.  It tells you nothing about the mass of the Earth under you.

So the way to do it is to find an object with known mass and something detectably orbiting it at a known radius.  Then G can be determined, and mass of Earth along with it.  Is that how it was done???  What object possibly fits that description?

[chris (OP)]

Thanks @Halc

[evan_au]

It seems easy to us now, but another piece of the puzzle in applying Newton's F = GmM/r² is to measure "r", the radial distance between astronomical bodies.

You can't exactly pace out the distance between the Earth and the Sun...

You can apply Newton's gravitational laws and Kepler's orbital laws to work out the relative sizes of the of the planetary orbits (the unknown mass of the Sun M being the same in all cases). But it doesn't give you the actual mass of the Earth unless you know G, M and r (Cavendish's laboratory measurement of G wasn't done until 1798).

In 1769, there was a transit of Venus (ie Venus passes between the Earth and the Sun).
- Observatories around the world recorded the exact time that Venus touched the edge of the Sun's disk, and where it touched the edges.
- One expedition under James Cook was sent to observe this event from Tahiti, in the South Pacific.
- By combining all these measurements they were able to work out the distance between Earth and Sun (ie "r") to within 1% of the distance calculated today using radar reflections from the planets.

On his way back to England, Cook explored New Zealand and the east coast of Australia. This event is taught in Australian schools, at it explains one encounter in our history.

See: https://en.wikipedia.org/wiki/1769_Transit_of_Venus_observed_from_Tahiti

[Bored chemist]

Cavendish's work refers to the process as measuring the density of, or weighing, the Earth.

[alancalverd]

from: alancalverd on 12/10/2018 09:49:23Using F = GmM/r2 you can calculate the force on a falling object of mass m in terms of M, the mass of the earth, and G, which we assume to be a universal constant.

That doesn’t work.  We’re trying to compute at least a rough G and M here.  We don’t know either of them yet.  We do know force F is 9.8 newtons for a 1KG mass.  We can assume we know r.  We therefore know the product of G and M, but not either separately.

We don't "know" F = 9.8 N/kg. We define the newton as1 kg.m.s^-2, and have a measured value of g = F/m

We don't "assume" r: it is a measured value - see below.

As I said, Cavendish (1797) measured G directly. 

Maskeleyne estimated the density of the planet from an experiment in 1774 at Schiehallion, measuring the deflection of a plumb bob on two sides of the mountain. In fact, that was a measure of G rather than ρ_earth since the assumption that the mountain was homogeneous was more defensible. Newton had considered the possibility but rejected the experiment as too difficult.

As F = ma, we can measure the acceleration of a falling object or the period of a pendulum to get a value for F/m

F=ma works (F 9.8 = 1 (mass) * 9.8 m/sec acceleration), but that doesn’t yield either mass of Earth M nor G, which are the two things we’re trying to determine here.
The pendulum thing is a function of acceleration (9., not of the mass of Earth.  Put a pendulum in a rocket accelerating at that rate and it will have the same period as here on Earth.  It tells you nothing about the mass of the Earth under you.

You have missed the point. If we know F/m we know GM/r², and we have known r since 200 BC (Eratosthenes), so all that was required for an accurate estimate of M was Cavendish's determination of G. The reason for using a pendulum  is that we know the period T = 2π√(Lm/F) for small oscillations, which is a simpler experiment than measuring the acceleration of an object in free fall.

The most amazing fact in all this, hundreds of years and thousands of experiments later,  is that the gravitational masses of all bodies are identical to their inertial masses. Why?

Interestingly, Hipparchus estimated the radius of the moon's orbit around 130 BC, though his assumptions regarding the earth's orbit of the sun were somewhat wild.

[alancalverd]

Cavendish's work refers to the process as measuring the density of, or weighing, the Earth.

That was his ultimate objective, but what his torsion balance experiment actually did was to measure G in terms of the force between known test masses.


Apologies for placing my note after BC's reply! Editorial cockup, not a wormhole in spacetime..