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

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.

Answer 1

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

Answer 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).

Answer 3

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.

Answer 4

c/ii anyone?

Answer 5

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 )