directly proportional to its distance from Earth imply about the universe?","back":"The universe is expanding (space is stretching in all directions).","points":1,"mcqOptions":[{"text":"The universe is expanding (space is stretching in all directions).","correct":true,"explanation":"A positive proportional relationship means increasingly distant galaxies recede increasingly quickly. This is evidence that space itself is stretching throughout the universe rather than galaxies moving from one central point."},{"text":"The universe is contracting (space is shrinking in all directions).","correct":false,"explanation":"Contraction would cause distant galaxies to approach us, producing negative recessional velocities or blueshifts. Positive recessional velocities increasing with distance instead show that the universe is expanding."},{"text":"The universe is static (space is unchanging in all directions).","correct":false,"explanation":"A static universe would not produce a systematic increase in recessional velocity with distance. The observed proportional relationship shows that space is expanding in all directions."},{"text":"The universe is rotating (space is turning in one direction).","correct":false,"explanation":"Rotation would produce direction-dependent motion, with some galaxies approaching and others receding. The general distance-recession relationship instead indicates that the universe is expanding as space stretches."}]},{"id":"85d18caf-3119-4b77-be2f-25b448442e65","cardType":"basic","front":"State the equation for Hubble's Law.","back":"$$v = Hd$","points":1,"mcqOptions":[{"text":"$$v = Hd$","correct":true,"explanation":"Hubble’s Law states that a galaxy’s recessional velocity $v$ equals the Hubble constant $H$ multiplied by its distance $d$, giving a directly proportional relationship."},{"text":"$$v = \\frac{H}{d}$","correct":false,"explanation":"Dividing by distance incorrectly makes recessional velocity decrease as distance increases. Hubble’s Law states that velocity is directly proportional to distance, so the correct equation is $v = Hd$."},{"text":"$$v = \\frac{d}{H}$","correct":false,"explanation":"Dividing distance by the Hubble constant is dimensionally incorrect for velocity. Multiplying distance by $H$ supplies the required velocity units, giving $v = Hd$."},{"text":"$$H = vd$","correct":false,"explanation":"Multiplying velocity and distance does not produce the units of the Hubble constant. Rearranging Hubble’s Law gives $H = v/d$, while its standard form is $v = Hd$."}]},{"id":"e966da83-5fd1-42a0-912b-b84020aa8389","cardType":"basic","front":"Identify the units typically used for $v$, $d$, and $H$ in Hubble's Law.","back":"

$v$: $\\text{km s}^{-1}$
$d$: $\\text{Mpc}$ (megaparsecs)
$H$: $\\text{km s}^{-1} \\text{ Mpc}^{-1}$

","points":3,"mcqOptions":[{"text":"

$v$: $\\text{km s}^{-1}$
$d$: $\\text{Mpc}$ (megaparsecs)
$H$: $\\text{km s}^{-1} \\text{ Mpc}^{-1}$

","correct":true,"explanation":"Recessional velocity is conventionally measured in kilometres per second and distance in megaparsecs. Therefore, $H = v/d$ has units of kilometres per second per megaparsec."},{"text":"

$v$: $\\text{km s}^{-1}$
$d$: $\\text{km}$ (kilometres)
$H$: $\\text{km s}^{-1} \\text{ Mpc}^{-1}$

","correct":false,"explanation":"Using kilometres for $d$ is inconsistent with the stated units of $H$, which are per megaparsec. The conventional units are $\\text{km s}^{-1}$ for $v$ and $\\text{Mpc}$ for $d$."},{"text":"

$v$: $\\text{m s}^{-1}$
$d$: $\\text{Mpc}$ (megaparsecs)
$H$: $\\text{km s}^{-1} \\text{ Mpc}^{-1}$

","correct":false,"explanation":"This mixes metres per second for $v$ with kilometres per second in $H$, introducing a factor-of-1000 inconsistency. Conventionally, $v$ uses $\\text{km s}^{-1}$ and $d$ uses $\\text{Mpc}$."},{"text":"

$v$: $\\text{km s}^{-1}$
$d$: $\\text{Mpc}$ (meg

Hubble's law (A-level only)

Hubble's law (A-level only) Flashcards

Hubble's law (A-level only)

Hubble's law (A-level only) Flashcards