Q67 / 79 · NEST (NISER) 2026 · Physics
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 $2 \times 10^{15}$ kg and $10$ km respectively. Take the mass of the Earth to be $6 \times 10^{24}$ kg. The gravitational acceleration (in SI units) of the Earth due to the asteroid just before the impact would be of the order
- A$10^{-5}$
- Correct answerB$10^{-9}$
- C$10^{-1}$
- D$10^{1}$
Solution
The gravitational force exerted by the asteroid on the Earth is given by Newton's law of gravitation:
$F = \dfrac{G M_a M_e}{r^2}$
where $M_a$ is the mass of the asteroid, $M_e$ is the mass of the Earth, and $r$ is the distance between their centers. Just before impact, $r = R_e + R_a \approx R_e$, where $R_e = 6.4 \times 10^6$ m is the radius of the Earth.
The acceleration of the Earth due to this force is:
$a_e = \dfrac{F}{M_e} = \dfrac{G M_a}{r^2}$
Substituting the standard value of $G$ and the given mass of the asteroid:
$a_e = \dfrac{6.67 \times 10^{-11} \times 2 \times 10^{15}}{(6.4 \times 10^6)^2}$
$a_e = \dfrac{1.334 \times 10^5}{4.096 \times 10^{13}} \approx 3.25 \times 10^{-9} \text{ m/s}^2$
The order of magnitude of the acceleration is $10^{-9}$.
$F = \dfrac{G M_a M_e}{r^2}$
where $M_a$ is the mass of the asteroid, $M_e$ is the mass of the Earth, and $r$ is the distance between their centers. Just before impact, $r = R_e + R_a \approx R_e$, where $R_e = 6.4 \times 10^6$ m is the radius of the Earth.
The acceleration of the Earth due to this force is:
$a_e = \dfrac{F}{M_e} = \dfrac{G M_a}{r^2}$
Substituting the standard value of $G$ and the given mass of the asteroid:
$a_e = \dfrac{6.67 \times 10^{-11} \times 2 \times 10^{15}}{(6.4 \times 10^6)^2}$
$a_e = \dfrac{1.334 \times 10^5}{4.096 \times 10^{13}} \approx 3.25 \times 10^{-9} \text{ m/s}^2$
The order of magnitude of the acceleration is $10^{-9}$.
Hence, the correct answer is option: 'B'.