Can someone just have a quick look over this question and my current answers and tell me if there are any massive mistakes im making. I feel confident with what ive done but just need a little boost.

Question:

An asteroid of mass m =7 .5×10^12 kg has two rocket thrusters attached to it. One thruster supplies a force with x- and y-components

F1 = (8 .0×10^7,7.0×10^7)N while the other supplies a force with x- and y-components
F2 = (4 .0×10^7,−2.0×10^7)N.
(You may ignore the mass of the rocket thrusters and their fuel.)

(a)
(i) What are the x- and y-components of the resultant force acting on the asteroid?

(ii) What is the magnitude of the resultant force and in which direction does it act?
(b) At what distance from the Earth, in kilometres, would the force calculated in (a) have the same magnitude as that due to the gravitational attraction between the asteroid and the Earth? Show that you have checked that your answers for the distance and its unit are sensible.

(c)
(i) What is the magnitude of the acceleration of the asteroid produced by the thrusters and in which direction does it act?
(ii) If the constant force calculated in (a) acts in the same direction as the asteroid’s motion, calculate the time required to change the speed of the asteroid by 1.0ms−1 and therefore show that this is about 16 hours.

My Attempt so far:

F1 = (8.0∗10^7N,7.0∗10^7N)
F2 = (4.0∗10^7N,−2.0∗10^7N)

For the resultant force we need F = F1 + F2.

a)
i)

F = (8.0 + 4.0,7.0−2.0)∗10^7N = (1.2∗10^8N,5.0∗10^7N)
|F| =sqrt((1.2∗10^8)^2 + (5.0∗10^7)^2)
|F| = 1.3∗10^8N

ii)
a = Fm
a = 1.3∗10^8N*7.5∗10^12kg
=1.73∗10^−5ms^−2
1 7.5∗10^12 ∗(1.2∗10^8N,5.0∗10^7N) = (1.6∗10^−5N,6.667∗10^−6N)ms^−2

b)
To find the distance from Earth where the magnitude is equal to the resultant force found above I will use F = Gm1m2/d^2 where d^2 is the variable I wish to isolate.

Rearranging this equation leaves us with:
d=sqrt(Gm1m2/F)

d=sqrt(6.67∗10^−11∗7.5∗10^12∗5.97∗10^24 1.3∗10^8)

d = 4.79302∗10^9m

So the asteroid would need to be around 4.8∗10^9 meters from the Earth for the force to have the same magnitude gravitational force as the force applied by the thrusters. To put this in to perspective this distance is around 2.9 million miles.

ii)

a = Fm

NO a = F/m but you seem to have divided ok anyway

a = 1.3∗10^8N*7.5∗10^12kg
=1.73∗10^−5ms^−2

What is this>?
1 7.5∗10^12 ∗(1.2∗10^8N,5.0∗10^7N) = (1.6∗10^−5N,6.667∗10^−6N)ms^−2

a.

i. The problem does NOT ask for the resultant force, it only asks for the x- and y-components of the resultant force:
x1+x2 = 12*10^7.
y1+y2 = 5*10^7
(x,y) = (12*10^7, 5*10^7).

a.

ii. Tan A = Y/X = 5*10^7/12*10^7 = 0.41667, A = 22.6o = The direction.

12*10^7/Cos22.6 = 13*10^7 = 1.3*10^8N. = Magnitude.

c/ii anyone?

a=∆v/∆t

You know what ∆v is and you know what a is.

rearrange to solve for ∆t = approx 57803 s ( about 16 hours )

Your attempt so far looks good! Here's a quick review of your answers:

(a)
(i) The x-component of the resultant force is correct at 1.2×10^8 N. However, the y-component should be -3×10^7 N instead of 5×10^7 N. So the correct answer is (1.2×10^8 N, -3×10^7 N).
(ii) The magnitude of the resultant force is correct at 1.3×10^8 N. The direction can be determined using trigonometry, where the angle θ = arctan(y-component/x-component). In this case, θ = arctan(-3×10^7 / 1.2×10^8) ≈ -14.1 degrees (assuming counterclockwise from the positive x-axis).

(b) The calculation and unit conversion for the distance are correct. The asteroid would need to be around 4.8×10^9 meters or 2.9 million miles from Earth.

(c)
(i) The magnitude of the acceleration is correct at 1.73×10^-5 m/s^2. The direction can be determined using the same logic as in part (a)(ii).
(ii) To calculate the time required to change the speed by 1.0 m/s, you need to use the equation v = u + at, where v is the final speed, u is the initial speed (assuming it's zero), a is the acceleration, and t is the time. Rearranging the equation, t = (v - u) / a. Plug in the values, t = (1.0 m/s - 0) / 1.73×10^-5 m/s^2 ≈ 57,803 seconds. This is approximately 16 hours (since there are 3600 seconds in an hour).

Overall, your answers are accurate. Well done!

Looking at your attempt, overall you seem to be on the right track. However, there are a few mistakes and areas that could use some improvement. Let's go through each part of your solution and make some corrections.

For part (a):

(i)
The addition of vectors F1 and F2 should be done component-wise, since they have both x- and y-components. Therefore, the correct addition should be:
F = (8.0×10^7 + 4.0×10^7, 7.0×10^7 - 2.0×10^7) N
F = (1.2×10^8 N, 5.0×10^7 N)

(ii)
To find the magnitude of the resultant force, you correctly used the formula for calculating the magnitude of a vector. However, you made a mistake in the calculation. The correct calculation should be:
|F| = sqrt((1.2×10^8)^2 + (5.0×10^7)^2)
|F| = 1.3×10^8 N

For part (b):

To find the distance from the Earth, you correctly used the formula for gravitational force. However, you made a mistake in calculating the distance. The correct calculation should be:
d = sqrt(G * m1 * m2 / F)
d = sqrt(6.67×10^-11 * 7.5×10^12 * 5.97×10^24 / 1.3×10^8)
d = 4.79×10^9 m

So the corrected distance is approximately 4.8×10^9 meters or 4.8 billion meters.

To check the correctness of your answer, it is important to consider whether the resulting distance is sensible. In this case, a distance of 4.8 billion meters, which is approximately 2.9 million miles, is reasonably consistent with the scale of astronomical distances.

For part (c):

(i)
To find the magnitude of the acceleration, you correctly used the formula a = F/m. However, you made a mistake in the calculations. The correct calculations should be:
a = F / m
a = (1.3×10^8 N) / (7.5×10^12 kg)
a = 1.73×10^-5 m/s^2

(ii)
To calculate the time required to change the speed of the asteroid by 1.0 m/s, you need the formula t = Δv / a. However, you did not provide the change in velocity (Δv). Without that, it is not possible to calculate the time required accurately.

To summarize:

- Make sure to perform vector addition component-wise when dealing with vectors with both x- and y-components.
- Review your calculations for magnitude and double-check for mistakes.
- Provide the necessary values for calculations, such as the change in velocity for calculating time.

Overall, your understanding of the concepts and approach to the problem appears to be good. By following these corrections, you should be able to get more accurate and sensible answers.