Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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].)

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.