Respuesta :
Answer:
[tex]P_f = (5.51\times 10^7 \hat i + 5.67 \times 10^7) kg m/s[/tex]
Explanation:
Force on Asteroid due to Earth's gravity is given as
[tex]F = \frac{Gm_1m_2}{r^2}\hat r[/tex]
now we know that
[tex]F = \frac{(6.67 \times 10^{-11})(1200)(5.98\times 10^{24})(9.00\times 10^6\hat i + 10.00\times 10^6\hat j}{((9.00\times 10^6)^2 + (10.00\times 10^6)^2)^{1.5}}[/tex]
[tex]F = 1769\hat i + 1965.5\hat j[/tex]
Now we can use impulse momentum theorem to find the final momentum of the Asteroid
[tex]P_f = P_i + F\Delta t[/tex]
[tex]P_f = 1200(4.30\times 10^4\hat i + 4.40\times 10^4\hat j) + (1769\hat i + 1965.5\hat j)(2.00\times 10^3)[/tex]
[tex]P_f = (5.51\times 10^7 \hat i + 5.67 \times 10^7) kg m/s[/tex]
The momentum of the asteroid is [tex]5.51 \times 10^7\hat {i} + 5.67 \times 10^7\hat {j} \;\rm kgm/s[/tex].
How do you calculate the momentum?
Given that the mass of the asteroid is 1200 kg and at a particular instant the asteroid’s velocity is ⟨ 4.30 x 104, 4.40 x 104, 0 ⟩ m/s, and its position with respect to the center of the Earth is ⟨ 9.00 x 106, 10.00 x 106, 0 ⟩ m. The center of the Earth (mass = 5.97 x 1024 kg), is at the origin of the coordinate system.
The force on the asteroid is given as below.
[tex]F = \dfrac {Gm_1m_2}{r^2} \hat{r}[/tex]
Where m1 is the mass of the asteroid and m2 is the mass of earth. G is the gravitational constant and r is the distance of the asteroid from the center of the earth.
[tex]F = \dfrac {6.67 \times 10^{-11} \times 1200 \times 5.97\times 10^{24}}{((9\times 10^6)^2 +(10\times 10^6)^2)^{1.5}}\times ( 9\times 10^6 \hat{i} + 10\times 10^6\hat {j})[/tex]
[tex]F = 1773 \hat {i} + 1970 \hat {j}[/tex][tex]P_f = 1200 ( 4.3 \times 10^4 \hat {i} + 4.4 \times 10^4 \hat{j}) + (1773 \hat {i} + 1970 \hat {j} ) \times (2 \times 10^3)[/tex]
The final momentum of the asteroid is calculated as given below.
[tex]P_f = P_1 + F \Delta t[/tex]
Where Pf is the final momentum, P1 is the initial momentum and F is the force at the asteroid.
[tex]P_f = 1200 ( 4.3 \times 10^4 \hat {i} + 4.4 \times 10^4 \hat{j}) + (1773 \hat {i} + 1970 \hat {j} ) \times (2 \times 10^3)[/tex]
[tex]P_f = 5.51 \times 10^7 \hat {i} + 5.67 \times 10^7 \hat {j} \;\rm kgm/s[/tex]
Hence we can conclude that the momentum of the asteroid is [tex]5.51 \times 10^7\hat {i} + 5.67 \times 10^7\hat {j} \;\rm kgm/s[/tex].
To know more about the momentum, follow the link given below.
https://brainly.com/question/904448.
