is it possible for you to explain it, kinda lost
Method 1 requires you knowing that the moons gravity is 1.2 ms^-2, but I doubt just stating it would get you the marks since its not explicitly stated in the syllabus. This one is more simpler than method 2.
This is what I did:
divide both sides by m
g = -1.62 approx (use actual value on calc).
considering the y axis/vertical direction only
v = 1158.79 ms^-1 (using rounded value of g) and so v = 1159 ms^-1.
Method 2 considers the energy within the system. This is more complex than method 1 but I'll give an idea whats going on
You can consider TE as mechanical energy, its U + K, I'll call it mechanical energy since thats more in line with the syllabus.
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The mechanical energy within the system is conserved so therefore;
This should be something familiar from Dynamics, try to think back to when you did momentum, you would try to find the final/initial velocity in a similar fashion.
The initial velocity is already given in the question so substitute that into initial KE. Mass is given as well. Remember that for final potential, the distance is different as it is the height of the mass (30km) + the radius of the moon (1740 km) so ensure to input that in [You'll see that in jimmy's working].
That's pretty much whats going on. I'd prefer method 1 but great to know other perspectives.
(d)
pretty much just using the orbital velocity formula.
When an object is in orbit of a body, it has a centripetal force like any object in circular motion. Centripetal force just refers to the force that acts radially inwards to another object, its not actually a force itself such as gravitational force. The centripetal force in this case would be the gravitational force and so the equation for gravitational force and centripetal force and equivalent. From this you can derive the velocity in which the object would be in orbit; hence orbital velocity.
and then just follow the working out from there