N (Whole number)

Since F = mg, and in this case m is 10kg and g is 9.8:

F = 10 x 9.8

F = 98 N

2. What is the GPE of a satellite orbiting a planet given:

     Mass of satellite: 1000kg

     Mass of Planet:     6 x 10¹⁶ kg

     Radius of Orbit:     3 x 10³ km

     Universal Gravitational Constant:    6.67 x 10^-11

J (Whole Number)

You will see from your formula sheet that the formula for GPE is:

E_p = (-Gm₁m₂) / r

From this, it is a simple substitution, the only trick being that you have to convert the 'Radius of Orbit' from kilometres to metres.

Therefore, the GPE is equal to 1334 J.

ms^-2 (Two decimal places)

For this we use the formula:

The only tricks being that we must convert the value for l into metres. Eg: 120cm = 1.2m.

We then must find the time for just one oscillation. This is simply 22.3 seconds divided by 10 (the number of oscillations).

From there we simply substitute into the formula so that we get:

g = (4 x ∏² x 1.2) / (2.23)²

Which gives g to be 9.53.

     Mass of Jupiter:    1.9 x 10²⁷ kg

     Radius of Jupiter:  7.15 x 10⁴ km

     Universal Gravitational constant:    6.67 x 10^-11

ms^-2 (Two decimal places)

This is solved using the formula:

The radius of Jupiter must first be converted to metres, but once this step is overcome the calculation is simple.

g = ((6.67 x 10^-11) x (1.9 x 10²⁷)) / (7.15 x 10⁷)²

From this, the value for g on the surface of Jupiter is 24.79 ms^-2.

N (Whole Number)

There is two parts to this question. The first of these is to determine my mass. Since F = ma, we have the following statement:

784 = 9.8m

From this, m = 80kg.

Now we use F = ma again to determine my weight force on Mars.

F = 80 x 3.7

Therefore, my weight force on Mars is 296 N.

ms^-1 (One decimal place )

This is a bit of a trick question. For it, we use the formula:

But, as we can see, there is no place in the formula for the mass of the satellite. The mass of the satellite is a useless fact designed to distract you. The only other tricky part to this question is remembering that the radius of the planet must be converted from kilometres to metres.

Once this is done, we have the expression:

v² = (2 x [6.67 x 10^-11] x [4 x 10²²]) / (6.134 x 10⁶)

v² = 869,905.4451

v = 932.7 ms^-1

ms^-2 (One decimal place )

Most people know that 'g' forces can be calculated by putting apparent weight as a fraction of true weight. What many tend to forget is that this expression can be re-written as:

Since we've been told to assume that the normal value for 'g' is 9.8, and that the maximum safe 'g' force is 8, we have an expression that reads:

8 = (9.8 + a) / 9.8

78.4 = 9.8 + a

68.6 = a

Which therefore means that the maximum acceleration an astronaut can handle is 68.6 ms^-2.

     Mass of Rocket (with fuel):    3.4 x 10⁴ kg

     Mass of Fuel:         170 kg

     Velocity of Gases being expelled:    5.6 x 10⁵ ms^-1

kmh^-1 (Whole Number)

This question is solely about momentum. The momentum of the rocket must equal the momentum of the gases.

The only real trick to the question is that when dealing with the mass of the rocket, we must remember to subtract the mass of the fuel as the question refers to the exact moment when the fuel runs out.

Therefore the mass of the rocket at this time is:

(3.4 x 10⁴) - 170 = 3.383 x 10⁴ kg

Now we simply substitute into the formula:

(3.383 x 10⁴) x v = 170 x (5.6 x 10⁵)

v = 2814.0704 ms^-1.

All that remains is to convert this figure to kilometres per hour by multiply by 3.6.

The velocity of the rocket is therefore 10131 kmh^-1.

N (One decimal place )

This is a pretty straightforward question. For this we use the formula:

The only trick to it is that we must remember to convert 65 kmh^-1 to metres per second by dividing by 3.6.

v = 65 / 3.6

From there it is a very simple substitution:

F = [1200 x (65 / 3.6)²] / 167

Which means that the centripetal force is 2342.5 N.

ms^-1 (One decimal place )

To find the orbital velocity of the satellite we use the formula which is derived from Kepler's Law of Periods.

Once again, the mass of the satellite itself is irrelevant, and the fact that it has been supplied is designed to be a distraction.

The other trick is that the radius of the orbit includes both the radius of the planet PLUS the altitude of the satellite's orbit. We must also ensure both figures are in metres.

For that reason, the radius of the orbit is equal to:

r = (6.134 x 10⁶) + (100 x 10³)

r = 6.234 x 10⁶ m

From there a simple substitution occurs:

v² = ([6.67 x 10^-11] x [6.7 x 10²¹]) / [6.234 x 10⁶]

v² = 71,685.916

v = 267.7 ms^-1.