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a sphere of mass 40kg is attracted by a sphere of mass 80kg with aforce equal to 1/4th of a milligram weight when their centres are 30cm apart . calculate value of G

Sagot :

jacob193

Answer:

Approximately [tex]6.898\times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex].

Explanation:

Between two spherical mass of uniform density, the gravitational force [tex]F[/tex] would be:

[tex]\displaystyle F = \frac{G\, m_{1}\, m_{2}}{r^{2}}[/tex],

Where:

  • [tex]G[/tex] is the gravitational constant,
  • [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the mass of the two spheres, and
  • [tex]r[/tex] is the distance between the center of the two spheres.

In this question, [tex]F[/tex], [tex]m_{1}[/tex], [tex]m_{2}[/tex], and [tex]r[/tex] are given. To find the value of [tex]G[/tex], rearrange the equation above:

[tex]\displaystyle G = \frac{F\, r^{2}}{m_{1}\, m_{2}}[/tex].

Before substituting in the values, make sure all quantities that are involved are in standard units.

  • Force should be in Newtons. The question stated the value of [tex]F[/tex] in terms of milligram ([tex]1\; {\rm mg} = 10^{-3}\; {\rm g} = 10^{-6}\; {\rm kg}[/tex]) weight. Assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex]:
    [tex]\begin{aligned} F &= m\, g \\ &= (0.25 \times 10^{-6}\; {\rm kg}) \times (9.81\; {\rm N\cdot kg^{-1}}) \\ &= 2.7025 \times 10^{-6}\; {\rm N}\end{aligned}[/tex].
  • Mass should be measured in kilograms.
  • Distance should be measured in meters:
    [tex]r = 30\; {\rm cm} = 0.30\; {\rm m}[/tex].

Substitute the values into the expression for [tex]G[/tex]:

[tex]\begin{aligned} G &= \frac{F\, r^{2}}{m_{1}\, m_{2}} \\ &= \frac{(2.7025\times 10^{-6}\; {\rm N})\, (0.30\; {\rm m})^{2}}{(40\; {\rm kg})\, (80\; {\rm kg})} \\ &\approx 6.898 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}\end{aligned}[/tex].

(The reference value of the gravitational constant is approximately [tex]6.7 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex].)

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