Calculating the mass of the Earth

  • Thread starter Thread starter ccapanni
  • Start date Start date
  • Tags Tags
    Earth Mass
AI Thread Summary
The discussion focuses on calculating the mass of the Earth using the gravitational constant and the distance to the Moon. The initial calculation yielded a mass of 5.04x10^34 kg, which is significantly larger than the known mass of Earth, approximately 5.97x10^24 kg. A key error identified was the failure to convert the lunar month into SI units, which is essential for accurate calculations. Additionally, it was pointed out that the relationship between the Earth and Moon's masses should be considered, suggesting that the result should be adjusted by multiplying by approximately 0.988. Correcting these issues will yield a more accurate estimate of Earth's mass.
ccapanni
Messages
1
Reaction score
0

Homework Statement



The distance between the centres’ of the Earth and Moon is estimated to be 3.84x10^8 m. If the lunar month is 27.3 days, calculate the approximate mass of the Earth. (Assume the gravitational constant G=6.67x10^-11 Nm2 kg-2)

Homework Equations


[/B]
T = 2pi / sqroot(G * Mearth) * r^3/2

The Attempt at a Solution



27.3 = 6.283 / sqroot(6.67x10^-11 * Mearth) * 3.84x10^3/2
27.3 * sqroot(Mearth) = (6.283 / 8.14 x10^-6) * 7.525x10^12
27.3 * sqroot(Mearth) = 771867.3 * 7.525x10^12
27.3 * sqroot(Mearth) = 5.81x10^18
sqroot(Mearth) = 5,81x10^18 / 27.3
sqroot(Mearth) = 2.13x10^17
Mearth = 4.53x10^34

Lunar Month / Earth Month = 30.4/27.3 = 1.114

Mass Earth = 4.53x10^34 * 1.114 = 5.04x10^34 kg

This is far too large a nuber for the mass of the earth, considering that it is in the region of Nx10^24 kg,
What am i doing wrong?
 
Physics news on Phys.org
Hi ccapani

You did not convert time units of the lunar month to conform with the SI units in which G is expressed.

ccapanni said:
Lunar Month / Earth Month = 30.4/27.3 = 1.114
This bit is irrelevant and shouldn't be included in the solution.
 
the Orbit Time using the center-to-center distance results in the TOTAL mass of the 2 objects.
Earth's Mass is about 81x Moon's mass, so (instead of multiplying by 1.114, that relates to orbit around the Sun)
you should make your answer "specific to Earth" : multiply it by 81/82 (=0.988)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top