The gravitation force on the satellite will be aiming toward the centre of earth so acceleration of the satellite will also be aiming toward the centre of earth.
NEET UG 2015 — Physics Mechanics
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,
Held on 30 Apr 2015 · Verified 9 Jul 2026.
the total mechanical energy of S varies periodically with time.
the linear momentum of S remains constant is magnitude.
the acceleration of S is always directed towards the centre of the earth.
the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.
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A bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
Consider a water tank shown in the figure. It has one wall at $x=L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\left(\theta_0 \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is:  $\left(\operatorname{take} \theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\right.$ is the acceleration due to gravity)
A physical quantity $P$ is related to four observations $a, b, c$ and $d$ as follows : $P=a^3 b^2 / c \sqrt{d}$ The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 2 \%$ and $4 \%$ respectively. The percentage error in the quantity $P$ is
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If $\mathrm{F}_{\mathrm{A}}$ and $\mathrm{F}_{\mathrm{B}}$ are the forces applied by the breaks on cars A and B , respectively, then the ratio $\mathrm{F}_{\mathrm{A}} / \mathrm{F}_{\mathrm{B}}$ is :
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