Question

1. a 560 - n woman stands on top of a very tall ladder so she is one earth radius above earths surface. how much does she weigh?

a. zero
b. 140 n
c. 280 n
d. 560 n
e. none of the above

1. suppose the gravitational force between two spheres is 30 n. if the magnitude of each mass doubles, what is the force between the masses?

a. 8 n
b. 15 n
c. 30 n
d. 60 n
e. 120 n

1. the force of gravity on you is greatest when you are standing

a. just below earths surface
b. on earths surface
c. just above earths surface

1. what event will produce the greatest increase in the gravitational force between the two masses?

a. doubling the large mass
b. doubling the distance between the masses
c. reducing the small mass by half
d. reducing the distance between the masses by half
problem
two satellites of equal mass are put 30. m apart. a gravitational force of 2.0×10⁻⁷ n acts between them. what is the mass of each satellite?

Explanation:

Step1: 寫出萬有引力公式

萬有引力公式為\(F = G\frac{m_1m_2}{r^2}\)，其中\(G = 6.67\times10^{- 11}\text{N}\cdot\text{m}^2/\text{kg}^2\)，\(m_1\)、\(m_2\)為兩物體質量，\(r\)為兩物體質心距離。
已知兩衛星質量相等，設為\(m\)，\(r = 30m\)，\(F=2.0\times 10^{-7}N\)。
將已知條件代入公式可得\(2.0\times 10^{-7}=6.67\times 10^{-11}\times\frac{m\times m}{30^{2}}\)。

Step2: 求解\(m\)

先對\(2.0\times 10^{-7}=6.67\times 10^{-11}\times\frac{m^{2}}{30^{2}}\)進行變形：
\(m^{2}=\frac{2.0\times 10^{-7}\times30^{2}}{6.67\times 10^{-11}}\)
\(m^{2}=\frac{2.0\times 10^{-7}\times900}{6.67\times 10^{-11}}\)
\(m^{2}=\frac{1.8\times 10^{-4}}{6.67\times 10^{-11}}\approx2.7\times10^{6}\)
再開平方得\(m=\sqrt{2.7\times 10^{6}}\approx1.64\times 10^{3}\text{kg}\)

Answer:

\(1.64\times 10^{3}\text{kg}\)