# HWS1 - 1-1 Calculate the mass m of a body that weighs 600...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1-1* Calculate the mass m of a body that weighs 600 lb at the surface of the earth. SOLUTION — 18.65 slug 1-4 Calculate the gravitational constant g, in SI units, for a location on the surface of the sun. SOLUTION Gm -11 30 z ._3 z é;§ZQLLQ___llll§29lllQ__l z 0_2741(103) m/SZ a 274 m/SZ Ans, [6.960(108)]2 1—6* The gravitational acceleration at the surface of Mars is 3.73 m/s? and the mass of Mars is 6.39(10 ) kg. Determine the radius of Mars. SOLUTION GmM 2 , FM f—"“"“T“""‘ -‘ //Gmu _ 6.673(10 11)(6.39)(102q) - / —————————§—7§———————m——— : 3.381(106) m a 3.38(106) m Ansg From Eq. (1—3): gm : r : M g” 1—8 Determine the gravitational force, in kilonewtons, exerted by the earth on the moon. SOLUTION H)(5.976H1024)(7.350H1022) [3.844(1OB)]2 1.984(1020) N = 1.984(10‘7) kN _ 6.673(10‘ 1-12 A satellite is placed in orbit 1.6(106) m aboxe the surface of the moon:_ If the mass of the satellite is 3.0(10 ) kg, determine the .gravitgﬁional force exerted on the satellite by the moon. SOLUTION _ 6 6 h + rm - 1.6(10 ) + 1.738(10 ) = 3.338(106) m G _ _ F : “3mm 2 6.673 10 ‘1 3.0 10‘)<7.350 1022 2 ——a——L_————LL———LL————*—~———LL————l r [3.338(106)]2 13.21(103) N a 13.21 kN 1-13* Determine the weight W of a satellite when it is in orbit 8500 miles above the surface of the earth if the satellite weighs 7600 lb at the surface. SOLUTION From Eq. (1-3): From Table 1—1: _ _ 2.090(107) ftv h = 2.090(107) + 8500(5280) = 6.578(107) ft 7600[2.090(107)12 [6.578(107)]2 Ans. 1—21 The first U.S. satellite, Explorer I, had a mass of approximately 1 slug. Determine the force exerted on the satellite by the earth at the low and high points of its orbit which were 175 mi and 2200 mi, respectively, above the surface of the earth. SOLUTION rL = re + hL = 2.090(107) + 175(5280) = 2.132(107) ft r“ = re + hH = 2.090(107) + 2200(5280) = 3.251(107) ft F = m g = 1(32.17) = 32.17 lb GmemS F3: 2 I‘ Z ‘2_ 2— FOrO — FHrH — FLrL — Gmem F r2 ? 2 FL = 02“ : QELliii§L999%ilg—il— = 25.51 lb 2 25.5 lb Ans. rL {2.182(10 )1 F r2 7 2 EH = —959 = §3433i13L9§9%ng—11~ = 13.296 lb 2 13.30 lb Ans. rH {3.251(10 )1 —————-—————-——_—_—_., 1-48* The elongation of a bar of uniform cross section subjected to an axial force is given by the equation 5 = 39 AE What are the dimensions of E if 5 and L are lengths, P is a force, and A is an area? ' SOLUTION ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

HWS1 - 1-1 Calculate the mass m of a body that weighs 600...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online