Solveeit Logo
Login

Question

Question: Acceleration due to gravity as a function of \(r\) is given by: A. \(\dfrac{4}{3}\pi Gr\left( {A -...

Acceleration due to gravity as a function of rr is given by:
A. 43πGr(A−Br)\dfrac{4}{3}\pi Gr\left( {A - Br} \right)
B. 4πGr(A−Br)4\pi Gr\left( {A - Br} \right)
C. 43πGr(A−34Br)\dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{4}Br} \right)
D. 43πGr(A−43Br)\dfrac{4}{3}\pi Gr\left( {A - \dfrac{4}{3}Br} \right)

Explanation

Solution

To solve this question, we will first need to find the mass of the volume enclosed in the radius rr. For this we will use the surface mass density and volume and obtain the mass by integration. After that, we will put this value of mass in the formula of gravitational acceleration to determine our required answer.

Formulas used:
ρ=A−Br\rho = A - Br
where, ρ\rho is the surface mass density, AA andBB are constants and rr is the radius.
g=GMr2g = \dfrac{{GM}}{{{r^2}}}
where, gg is the acceleration due to gravity, GG is the gravitational constant, MM is the mass and rr is the radius.

Complete step by step answer:
Our first step is to find the mass of the volume enclosed in the radius rrwhich is given by:
M=∫0rρdVM = \int\limits_0^r {\rho dV}
We can write ρ=A−Br\rho = A - Br and dV=4πr2drdV = 4\pi {r^2}dr.
M=∫0r(A−Br)4πr2dr ⇒M=4π∫0r(Ar2−Br3)dr ⇒M=4π(Ar33−Br44) ⇒M=43πr3(A−34Br) M = \int\limits_0^r {\left( {A - Br} \right)4\pi {r^2}} dr \\\ \Rightarrow M = 4\pi \int\limits_0^r {\left( {A{r^2} - B{r^3}} \right)} dr \\\ \Rightarrow M = 4\pi \left( {A\dfrac{{{r^3}}}{3} - B\dfrac{{{r^4}}}{4}} \right) \\\ \Rightarrow M = \dfrac{4}{3}\pi {r^3}\left( {A - \dfrac{3}{4}Br} \right) \\\
Now we will put this value in the formula of acceleration due to gravity.
g=GMr2 ⇒g=Gr243(πr3(A−34Br)) ∴g=43πGr(A−34Br) g = \dfrac{{GM}}{{{r^2}}} \\\ \Rightarrow g = \dfrac{G}{{{r^2}}}\dfrac{4}{3}\left( {\pi {r^3}\left( {A - \dfrac{3}{4}Br} \right)} \right) \\\ \therefore g = \dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{4}Br} \right) \\\
Thus, we can say that the acceleration due to gravity as a function of rr is given by 43πGr(A−34Br)\dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{4}Br} \right).

Hence, option C is the right answer.

Note: While doing the integration we have taken dV=4πr2drdV = 4\pi {r^2}dr. This is because we have considered a very small strip of the sphere having radius drdr. Therefore, its volume can be given by the multiplication of the surface area of the sphere to its radius. The surface area of this strip or element will be 4πr24\pi {r^2}. Thus we have taken the small element and then by integration, we have obtained the mass enclosed in the radius rr.