Homework #7-solutions

Homework #7-solutions - nichols (scn285) Homework #7...

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nichols (scn285) – Homework #7 – Erskine – (58200) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Newton’s law oF universal gravitation is F = G Mm r 2 . Here, M and m are masses and r is the sep- aration distance. The dimension oF Force is specifed by the equation F = ma . What are the SI units oF the constant G ? 1. [ G ] = kg / m 2 / s 2 2. [ G ] = N m 3. [ G ] = m 3 / kg 2 / s 2 4. [ G ] = N m / s 2 5. [ G ] = m / kg / s 2 6. [ G ] = m 3 / kg / s 2 correct 7. [ G ] = W / m 3 8. [ G ] = m 2 / kg 2 / s 2 9. [ G ] = m 2 / kg 10. [ G ] = J s / kg Explanation: F = G r 2 so G = F r 2 . Dimensional analysis oF G : ± kg m s 2 ² m 2 (kg) 2 = m 3 (kg s 2 ) . The common expression is [ G ] = m 3 / kg / s 2 . 002 (part 1 oF 7) 10.0 points G is the universal gravitational con- stant G =6 . 67 × 10 - 11 Nm 2 /kg 2 g is the Free Fall acceleration at Earth’s sur- Face g =9 . 81 m/s 2 . R E is the mean radius oF the Earth M E is the mass oF the Earth. Basics At the Earth’s surFace, the gravi- tational attraction oF the Earth on a body oF mass m can be stated in terms oF either g or G . Let ˆ r be a unit vector, and $g = g ˆ r ; the correct Formulae are 1. $ F g = - m$g and $ F G = - m GM E R 2 E $ R E . 2. $ F g = and $ F G = - m E R 2 E ˆ r. 3. $ F g = - and $ F G = - m E R 2 E ˆ cor- rect 4. $ F g = and $ F G = - m E R 2 E $ R E . Explanation: $ F g = - and $ F G = - m E R 2 E ˆ r 003 (part 2 oF 7) 10.0 points Why is $ F g ∝- ˆ r ? 1. There are no negative masses. 2. Gravitational Forces can be either attrac- tive or repulsive. 3. Gravitational Forces are always repul- sive. 4. Gravitational Forces are always attractive. correct Explanation: This is because the gravitational Forces are attractive. 004 (part 3 oF 7) 10.0 points The physical dimensions oF G and g are 1. [ G ]= L - 2 M - 1 T - 2 and [ g L 3 M - 1 T - 2 .
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nichols (scn285) – Homework #7 – Erskine – (58200) 2 2. [ G ]= LT - 2 and [ g - 2 . 3. [ G L 2 M - 1 and [ g M L T - 2 . 4. [ G L 3 M - 1 T - 2 and [ g - 2 . correct Explanation: [ G L 3 M - 1 T - 2 and [ g - 2 . 005 (part 4 of 7) 10.0 points On a planet where the mass density is the same as the Earth density but the radius R P is larger than R E , the gravitation Feld 1. g P >g E . correct 2. g P = g E . 3. cannot be determined because I do not know the ration of the masses M P M E of the planets. 4. g P <g E . Explanation: Equal densities means M P R 3 P = M E R 3 E . g P = GM P R 2 P . Replace M P by M E R 3 P R 3 E and E R 2 E g E . Thus g P = E R 2 E R P R E = g E R P R E E . 006 (part 5 of 7) 10.0 points Only potential energy di±erences are mean- ingful; therefore, when we talk about poten- tial, we must give a reference value for the potential. Let U o be the gravitational set equal to zero at the surface of the Earth; let U be the gravitational potential set equal to zero inFnitely far from the earth. The values of U o and U at an altitude h are 1. U o = - mg h and U = - mGM E R E + h 2. U o = and U = - m E R E h 3. U o = and U = - m E R 2 E h 4. U o = and U = - E R E + h cor- rect Explanation: U o = and U = - m E R E + h (check U o ( h = 0) = 0 ,U ( h = ) = 0).
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This note was uploaded on 04/06/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework #7-solutions - nichols (scn285) Homework #7...

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