Question

PLEASE SHOW ALL WORK. THANKS.

III, SIMPLE HARMONIC HECK (30 pts: 10 pts each piece), The statements immediately to follow (even when long and complicated) are considered in this context) GIVEN You may assume and rely on them for the problem/proof to follow a bit further down. Note: In some cases, "GIVEN' might mean 'self-evident' or 'obvious', but in other cases, it might not. GIVEN might not mean "obvious"; it can simply mean 'somehow established prior to this discussion'. GIVEN (for this context) → 1) Planet Earth can be treated as one large SOLID SPHERE of UNIFORM (constant) DENSITY, with constant ("known') mass M. constant ('known') radius R. 2) Given ANY TWO POINT MASSES. say mi and m2. then my will necessarily exert a gravitational force onto me and thereby pull me directly toward m. ("If we wish to be careful and precise in keeping track of directions, we can do the following: Let T be a displacement vector that points directly from my to m2 and is precisely as long as the distance between m1 and 12. then the customary scalar quantity referring to distance between masses. can be understood as the pure magnitude of the vector r. while the possibly less familiar r is understood as the pure direction of r.) Hence, the gravitational pull exerted by my onto my is given by force given by: F=- Gmim, ,2 r This Claim is symmetric (applies identically to the pull exerted by m2 onto m1). but is always and only a statement about the interaction which arisos BETWEEN 2 (infinitosimally small) POINTS OF MASS('particles'). Page 6 of 8

Answer 1

Solution :-

Solution to this problem is based on the concept of newtons law of garvitation and shell theorem. We also require knowledge of SHM, like second order differential equation of simple harmonic motion. But our required answer is T/2, because we are asked to calculate time required to reach from one side to other side of tunnel. Plesae read solution carefully and feel free to ask any doubt you face while understanding my solution. I would be happy to help.

(for) Force on of Region 2 R a sphere of mass M with 94 of two regions. s Region 1 Solno (a) a point mass m located within the volume enclosed by the sphere Mass M. Consider a point mass in lie at a distance er from the centre, inside uniform mass distribution (s (mass density) is Constant). We can divide the mass M solid sphere in One overlaying mass im is region I and the solid portion under laying is region 2 Invagine the overlayering portion to be composed of concentric uniform spherical shells. Each individual shell gravitational force since inside each Shell (Recall Shell theorem). so, The mass m, of the sphere in region 1 contributes zero- gravitational force outside the spherical mass (m) of region 2. So, mass of this region can be considered to located at the centre of the sphere for purpose of Calculating the gravitational force on m. Thus the magnitude of total gravitational face of the entire Sphine of mass M on m. is due to only portion of the total mass of the ephere (M) that lies in Zero total exerts m is on mass m, total on m. Mass m 18 be that Region 2.

As mass M is uniform so, Mass density (p) i.e. mass per unit volume = M MARS to mass me of ephese M in Region 2 is Ру 4 кон mi So, the magnitude of total gravitational force of the sphere on m is the gravitational attraction between mass in and mass me of sphere of mass M located at the centre of sphere Fiel G m m ê [where å is unit vector) fait G me mu 어 YARI GM me ere R3 912 flas/= Gemma b) a het Rm to be radius of moon. initial velocity of mass m, droped in the tunnel to be zero. * het a distance or from the centre of moon of mass M, is our Rm M mass

being the till calculaled due to moon м M is. of mass free --G Mm M (ve represents attractive nature of fora? from Newtons second law of motion & fama substitute Freima in eq. . Gмм 91 ма - der R3 where a =d? GM M=0 + dt? R² So mass m This is a second order differential equation of S.HM ( simple Harmonic Motion] droped inside tunnel will perform SHM. W2 GM and W = 24 T 4T²R3 GM R? 47² GM => T2 T2 R3 - or T= 47² R² is the time period. GM M = 7 X 10' 7x 100 kg 9 RE 1.78 loom for mass incide earth, the time period T can be calculated as a 4x3.14 x 1.7x107x1.7 x 108 61.76728x107 6.67810-h7 7X10224 46.69 3635.4340017 de £ 6o ob minuter m १ TE T= 24840.950062 6.8330081223 Required time = = 8o. 3 minutes

OVA 910 F m Rm C) Now, assume the particle of mass in 2 tunnel is at a distance & from the centre of tunnel and to being peependicular distance of tunned from centre of earth planet. Force on m • GMm Ve²+x² RS Mixe Force acting on the particle towards the center of the tunnel will behaves as force. Pour F sino GM m Fe²+2 Sino R3 .- GMm Sektu?.. sino 2 rest ſe2+x² R3 | H ma =-GM m x R3 dx GM x=0 + -dz dta R3 2 ام SHM Differential eq. of 4A² W = 20 T W 2 GM & GM R3 2 T RI T² 40?R? T- 4 T² R3 GM Time period is unchanged and is independent of how the tunnel ů dug.