iGCSE Physics (0625) 6.2.3 The Universe-Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed solution:
The Hubble constant $H_0$ is defined by the equation $H_0 = \frac{v}{d}$, which corresponds to the gradient of the speed-distance graph.
Selecting a point on the best-fit line, such as $d = 1.0 \times 10^{25} \text{ m}$ and $v = 20,000,000 \text{ m/s} = 2.0 \times 10^7 \text{ m/s}$.
Substituting these values into the formula: $H_0 = \frac{2.0 \times 10^7 \text{ m/s}}{1.0 \times 10^{25} \text{ m}}$.
This calculation yields $H_0 = 2.0 \times 10^{-18} \text{ s}^{-1}$.
Note that the units for the Hubble constant are $\text{s}^{-1}$ because the meters cancel out in the ratio of speed to distance.
Therefore, Option A is the correct value based on the gradient of the provided graph.
▶️ Answer/Explanation
Detailed solution:
The age of the Universe is estimated using the formula $t = \frac{d}{v}$.
The distance $d$ in km is $(60 \times 10^{6}\text{ light-years}) \times (9.5 \times 10^{12}\text{ km/light-year})$.
The recession speed $v$ is $1300\text{ km/s}$. Dividing $d$ by $v$ gives the age in seconds.
To convert the age from seconds to years, we divide the result by $3.2 \times 10^{7}\text{ s/year}$.
The final expression is $\text{Age} = \frac{60 \times 10^{6} \times 9.5 \times 10^{12}}{1300 \times 3.2 \times 10^{7}}$, which matches Option B.
▶️ Answer/Explanation
Detailed solution:
To find the distance $d$, we use the Hubble Law equation: $v = H_{0}d$, which rearranges to $d = \frac{v}{H_{0}}$.
The velocity $v$ is given as $4.4 \times 10^{6} \text{ m/s}$.
According to the syllabus, the current estimate for the Hubble constant is $H_{0} \approx 2.2 \times 10^{-18} \text{ s}^{-1}$.
Substituting the values: $d = \frac{4.4 \times 10^{6}}{2.2 \times 10^{-18}}$.
This calculation yields $d = 2.0 \times 10^{24} \text{ m}$.
Therefore, the approximate distance to the galaxy is $2 \times 10^{24} \text{ m}$, matching option C.