Question:
Physicist Luis Alvarez and his collaborators hypothesised that the extinction of dinosaurs was due to the impact of an asteroid with the Earth. They estimated the mass and the radius of the asteroid to be about $2 \times 10^{15}\text{ kg}$ and $10\text{ km}$ respectively. Take the mass of the Earth to be $6 \times 10^{24}\text{ kg}$. The gravitational acceleration (in SI units) of the Earth due to the asteroid just before the impact would be of the order
Physicist Luis Alvarez and his collaborators hypothesised that the extinction of dinosaurs was due to the impact of an asteroid with the Earth. They estimated the mass and the radius of the asteroid to be about $2 \times 10^{15}\text{ kg}$ and $10\text{ km}$ respectively. Take the mass of the Earth to be $6 \times 10^{24}\text{ kg}$. The gravitational acceleration (in SI units) of the Earth due to the asteroid just before the impact would be of the order
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Be careful whose acceleration is being asked!
We want the acceleration of the Earth due to the asteroid, so we must use the mass of the asteroid $M_a$ in the formula, not the mass of the Earth.
We want the acceleration of the Earth due to the asteroid, so we must use the mass of the asteroid $M_a$ in the formula, not the mass of the Earth.
Updated On: Jun 16, 2026
- $10^{-9}$
- $10^{1}$
- $10^{-1}$
- $10^{-5}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the gravitational acceleration experienced by the Earth due to the gravitational pull of the asteroid just before they collide.
Step 2: Key Formula or Approach:
The gravitational acceleration $a$ experienced by an object due to another body of mass $M$ at a distance $r$ is:
\[ a = \frac{G M}{r^2} \]
where $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$.
Just before impact, the distance between the center of the Earth and the center of the asteroid is approximately equal to the radius of the Earth $R_E \approx 6.4 \times 10^6\text{ m}$.
Step 3: Detailed Explanation:
• The mass of the asteroid is $M_a = 2 \times 10^{15}\text{ kg}$.
• The distance $r$ is the distance between their centers of mass just before impact. Since the Earth's radius ($6400\text{ km}$) is much larger than the asteroid's radius ($10\text{ km}$), we can approximate $r \approx R_E \approx 6.4 \times 10^6\text{ m}$.
• The gravitational acceleration of the Earth towards the asteroid is:
\[ a = \frac{G M_a}{R_E^2} \]
• Substituting the values:
\[ a = \frac{(6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2) \times (2 \times 10^{15}\text{ kg})}{(6.4 \times 10^6\text{ m})^2} \]
\[ a = \frac{1.334 \times 10^5}{4.096 \times 10^{13}} \]
\[ a \approx 3.26 \times 10^{-9}\text{ m/s}^2 \]
• The order of magnitude of $3.26 \times 10^{-9}$ is $10^{-9}$.
Step 4: Final Answer:
The gravitational acceleration is of the order of $10^{-9}$.
The question asks for the gravitational acceleration experienced by the Earth due to the gravitational pull of the asteroid just before they collide.
Step 2: Key Formula or Approach:
The gravitational acceleration $a$ experienced by an object due to another body of mass $M$ at a distance $r$ is:
\[ a = \frac{G M}{r^2} \]
where $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$.
Just before impact, the distance between the center of the Earth and the center of the asteroid is approximately equal to the radius of the Earth $R_E \approx 6.4 \times 10^6\text{ m}$.
Step 3: Detailed Explanation:
• The mass of the asteroid is $M_a = 2 \times 10^{15}\text{ kg}$.
• The distance $r$ is the distance between their centers of mass just before impact. Since the Earth's radius ($6400\text{ km}$) is much larger than the asteroid's radius ($10\text{ km}$), we can approximate $r \approx R_E \approx 6.4 \times 10^6\text{ m}$.
• The gravitational acceleration of the Earth towards the asteroid is:
\[ a = \frac{G M_a}{R_E^2} \]
• Substituting the values:
\[ a = \frac{(6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2) \times (2 \times 10^{15}\text{ kg})}{(6.4 \times 10^6\text{ m})^2} \]
\[ a = \frac{1.334 \times 10^5}{4.096 \times 10^{13}} \]
\[ a \approx 3.26 \times 10^{-9}\text{ m/s}^2 \]
• The order of magnitude of $3.26 \times 10^{-9}$ is $10^{-9}$.
Step 4: Final Answer:
The gravitational acceleration is of the order of $10^{-9}$.
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