The relative uncertainty in the period of a s | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

From Kepler's law, time period of a satellite,

$T = 2\pi \sqrt {\frac{{{r^3}}}{{Gm}}} \,\,\,\,\,\,\,{T^2} = \frac{{4{\pi ^2}}}{{GM}}{r^3}$

R

Vedclass · Accepted Answer

$2	imes 10^{-2}$

Vedclass · Answer

From Kepler's law, time period of a satellite,

$T = 2\pi \sqrt {\frac{{{r^3}}}{{Gm}}} \,\,\,\,\,\,\,{T^2} = \frac{{4{\pi ^2}}}{{GM}}{r^3}$

Relative uncertainty in the mass of the earth 

$\left| {\frac{{\Delta M}}{M}} ight| = 2\frac{{\Delta T}}{T} = 2 	imes {10^{ - 2}}$

$\left( {4\pi \ G constant and relative uncertainty

in radius \frac{{\Delta r}}{r}} negligible ight)$

The relative uncertainty in the period of a sateilite orbiting around the earth is $10^{-2}$. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is

Similar Questions

Kepler discovered

According to Kepler, the period of revolution of a planet $(T)$ and its mean distance from the sun $(r)$ are related by the equation

A binary star system consists of two stars one of which has double the mass of the other. The stars rotate about their common centre of mass :-

A geostationary satellite is orbiting the earth at a height $5R$ above the surface of the earth , $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is

If satellite is shifted towards the earth. Then time period of satellite will be