Help me with my homework. YOMAMA

[Gawdzilla Sama]

[klr]

I would so love to see that on an exam paper.

[Azathoth]

paging jim

[Calilasseia]

It would help if I could see the full text of the questions other than question 1. They look nice and easy.

[Gawdzilla Sama]

It would help if I could see the full text of the questions other than quhestion 1. They look nice and easy.

I think the missing word is "if".

[Calilasseia]

Ah, of course ... "open image in new tab". Use brain.

Question 2

Question 2 is nice and easy. Using the standard formula for universal gravitation:

F=GMm/r2

where M is the mass of the larger, gravitating object, and m is the mass of the attracted, accelerating object, r is the distance between the two objects, and G (the value of which is supplied in the question) is Newton's Constant of Universal Gravitation.

Allying this with the standard expression of Newton's Second Law, namely F=ma, we have:

a=GM/r2

We are interested in obtaining M, the mass of Yo Mamma. Algebraic rearrangement gives:

M = ar2/G

Therefore, feeding in those values, we have:

M = 3.74 × 1011 Kg.

Incidentally, if Yo Mamma really was that fat, either she'd be the size of a small asteroid, or she'd be so materially dense, that she'd be in danger of becoming a black hole.

Question 3

Meanwhile, question 3 assumes that we're dealing with circular motion. Under such motion, the standard relation is:

F = mv2/r

where m is the mass of the object in motion, v is the tangential speed of the object, and r is the radius of the circular path. Equating this with F = GMm/r2, we get:

mv2/r = GMm/r2

Cancelling appropriate terms on both sides, we get:

v2 = GM/r

Note that here, r is the distance from the centre of the Earth, toward which the gravitational force is acting.

Rearranging this gives us:

r = GM/v2

In addition, since the flight path is circular, the length of the flight path is 2πr, and therefore the time of one orbital revolution is given by:

T = 2πr/v

This gives us v = 2πr/T. Substituting this into the above expression for v gives us:

r = GMT2 ÷ 4π2r2, or

r3 = GMT2/4π2

This gives us r = 7.1 × 106 m. Feeding that into the expression:

v = 2πr/T

gives us v = 7.4 × 103 m s-1.

As a quick check, the value for r puts the object in Earth orbit at around 763 Km (474 miles) above the Earth's surface, and the orbital speed is around 16,700 mph. These figures are reasonable figures corresponding to those obtained from real low Earth orbit spaceflights.

Enjoy.

[Gawdzilla Sama]

No clue what all that meaqns.

[Calilasseia]

In which case, if this is an actual exam paper you're trying to pass, you're fucked.

[klr]

In which case, if this is an actual exam paper you're trying to pass, you're fucked.

That's my kind of exam paper.

Amongst other things, I run the exam papers system at our university, from the point the academics first submit them to central authority to when they get released for public consumption after the exams take place. Some day, I must save all those papers to an external drive for my own use ... I'll search through them for anything "interesting".

[Bella Fortuna]

I always wanted this book:

[Gawdzilla Sama]

In which case, if this is an actual exam paper you're trying to pass, you're fucked.

No shit, Sherlock.