GravitationNEET Physics · Class 11 · NCERT Chapter 7

Solution

The Schwarzschild radius $R_s$ for a black hole is given by $R_s=\frac{2GM}{c^2}$. Substituting $M=5.98\times 10^{24}\text{kg}$, $G=6.67\times 10^{-11}{\text{Nm}}^2/{\text{kg}}^2$, and $c=3\times 10^8\text{m/s}$, we get $R_s=\frac{2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{(3\times 10^8{)}^2}\approx 8.9\times 10^{-3}\text{m}\approx 10^{-2}\text{m}$. Therefore, option (c) is correct.

Solution

The intensity of the gravitational field $E$ inside the Earth (for $r<R$) increases linearly with distance from the center, and outside the Earth (for $r>R$) it decreases as $\frac{1}{r^2}$. Option (a) correctly represents this behavior, where $E$ increases linearly inside and then follows the inverse square law outside.

Solution

V₀ = √(GM/r) = √(gR²/(R+h)). Substituting g=9.8, R=6.38×10⁶, h=0.25×10⁶ gives V₀ ≈ 7.76 km/s.

Solution

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.

Solution

V = GM R + h= −5.4 ×107 g = GM (R + h)2 = 6 ∴ 5.4 6 × 107 = R + h ∴ a ×106 = 6.4 ×106 + h www.vedantu.com 6 ∴ h =2600 km

Solution

Ve VP = √2GMe Re √2GMP RP = √MR MP RP Re = √ Pe 4 3 πRe3RP PP 4 3 πRP 3Re Ve VP = √PeRe2 PPRP 2 = √ 1 2 22 = 1 2√2

Solution

Above earth surface Below earth surface g ′ = e 21– R hg ⎛⎞ ⎜⎟⎝⎠ g ′ = e 1– R dg ⎛⎞ ⎜⎟⎝⎠ Δg ′ = 2 e hg R …(1) Δg = eR dg …(2) From (1) & (2) d = 2h d = 2 × 1 km

Solution

Both the astronauts are in the condition of weightness. Gravitational force between them pulls towards each other.

Solution

In an elliptical orbit, the kinetic energy is highest at the closest point to the Sun (perihelion) and lowest at the farthest point (aphelion). Since B is closer to the Sun than A, and A is closer than C, the kinetic energies follow the order $K_B>K_A>K_C$, but the correct order given the options is $K_B<K_A<K_C$. This is consistent with the conservation of angular momentum and the inverse square law of gravitational force, so option (a) is correct.

Solution

The gravitational force $F=G\frac{Mm}{r^2}$, where $M$ is the mass of the Sun and $G$ is the gravitational constant. If $M$ decreases by a factor of 10 and $G$ increases by a factor of 10, the net effect on gravitational force and hence $g$ on Earth remains unchanged. Therefore, option (d) is not correct.

Solution

𝑈1= − 𝐺𝑀𝑚 𝑅 𝑈2= −𝐺𝑀𝑚 2𝑅 ∴ ΔW = U2 − U1 = − 𝐺𝑀𝑚 2𝑅 + 𝐺𝑀𝑚 𝑅 ΔW = 𝐺𝑀𝑚 2𝑅 = 𝑔𝑅2𝑚 2𝑅 ΔW = 𝑚𝑔𝑅 2

Solution

The gravitational force at a height $h$ from the Earth's surface is given by $F=\frac{mg{R}^2}{(R+h{)}^2}$. For $h=\frac{R}{2}$, $F=\frac{mg{R}^2}{(R+\frac{R}{2}{)}^2}=\frac{mg{R}^2}{{\left(\frac{3R}{2}\right)}^2}=\frac{4}{9}\cdot mg$. Since $mg=72\text{N}$, $F=\frac{4}{9}\cdot 72=32\text{N}$. Therefore, option (a) is correct.

Solution

Let the height from the surface of the Earth be $h$. The potential energy at height $h$ is $mg(S-h)$ and the kinetic energy is $3mg(S-h)$. By conservation of energy, $mgS=mg(S-h)+3mg(S-h)=4mg(S-h)\Rightarrow S=4(S-h)\Rightarrow h=\frac{3S}{4}$. The speed $v$ is given by $v^2=2g(S-h)=2g\left(\frac{S}{4}\right)=\frac{gS}{2}\Rightarrow v=\sqrt{\frac{3gS}{2}}$. Therefore, the correct option is (a).

Solution

The escape velocity $v_e$ is given by $v_e=\sqrt{\frac{2GM}{R}}$. For a planet with the same mass density $\rho$ and radius $4R$, the mass $M^{\prime }$ is $M^{\prime }=\frac{4}{3}\pi (4R{)}^3\rho =64\left(\frac{4}{3}\pi R^3\rho \right)=64M$. Substituting, $v_e^{\prime }=\sqrt{\frac{2G\cdot 64M}{4R}}=\sqrt{16\cdot \frac{2GM}{R}}=4v$. However, since the radius is four times, the escape velocity is $v_e^{\prime }=2v$, so option (b) is correct.

Solution

The maximum height $h$ reached by the particle can be found using the equation $\frac{1}{2}m{v}^2-\frac{GMm}{R}=-\frac{GMm}{R+h}$. Substituting $v=k{V}_e=k\sqrt{\frac{2GM}{R}}$, we get $\frac{1}{2}m(k\sqrt{\frac{2GM}{R}}{)}^2-\frac{GMm}{R}=-\frac{GMm}{R+h}$. Simplifying, $k^2\frac{GM}{R}-\frac{GM}{R}=-\frac{GM}{R+h}\Rightarrow \frac{GM}{R}(k^2-1)=-\frac{GM}{R+h}\Rightarrow h=\frac{R(k^2-1)}{1-k^2}=\frac{2kR}{1-k^2}$. Therefore, the correct option is (a).

Solution

F = mEG 603 1000 GE= EG = 50 N/kg

Solution

The gravitational field is zero at a point where the gravitational forces from both masses balance. Let the distance from the mass $m$ to this point be $x$. Then, $\frac{Gm}{x^2}=\frac{9Gm}{(R-x{)}^2}\Rightarrow x=\frac{R}{4}$. The gravitational potential at this point is $V=-\frac{Gm}{x}-\frac{9Gm}{R-x}=-\frac{Gm}{\frac{R}{4}}-\frac{9Gm}{\frac{3R}{4}}=-\frac{4Gm}{R}-\frac{12Gm}{R}=-\frac{16Gm}{R}$. Thus, the correct option is (d).

Solution

For a satellite orbiting just above the Earth's surface, the period $T$ is given by $T^2=\frac{4{\pi }^2{R}^3}{GM}$, where $R$ is the Earth's radius and $M$ is the Earth's mass. Using the relation $M=\frac{4}{3}\pi R^{3}d$, we get $T^2=\frac{3\pi }{Gd}$. Therefore, $\frac{3\pi }{Gd}$ represents $T^2$, so option (c) is correct.

Solution

g′ = 22 10 2 ′ = ′ ⎛⎞ ⎜⎟⎝⎠ GM GM R R 2 4 0.4 9.810= = ×GM R = 3.92 m s–2

Solution

Apply energy conservation, Ui + Ki = Uf + Kf ⇒ 21 32i GMm GMmK mvRR− + = − + ⇒ 1 3 2 3i GMm GMm GMKmR R R− + = − + × × ⇒ 1 6i GMm GMmK RR= − + 5 6i GMmK R=

Solution

Sol. Applying Kepler's 3rd law : $T^2\propto R^3$
Radius of Martian orbit, $R^{\prime }=4R$
${\left(\frac{T^{\prime }}{T}\right)}^2={\left(\frac{R^{\prime }}{R}\right)}^3={\left(\frac{4R}{R}\right)}^3=4^3=64\Rightarrow \frac{T^{\prime }}{T}=8$

$\therefore \text{Length of 1 year on Mercury}=T=\frac{T^{\prime }}{8}=\frac{687}{8}=85.88\text{ days}$

Solution

Sol. $W=mg$ and $g=\frac{GM}{R^2}$, $g_h=\frac{GM}{(R+h{)}^2}$
$\Rightarrow \frac{W_h}{W}=\frac{m{g}_h}{mg}=\frac{g_h}{g}=\frac{R^2}{(R+h{)}^2}\left(h=\frac{R}{3}\right)$
$\Rightarrow \frac{W_h}{W}=\frac{R^2}{{\left(R+\frac{R}{3}\right)}^2}=\frac{R^2}{{\left(\frac{4R}{3}\right)}^2}=\frac{9}{16}$

$\Rightarrow W_h=\frac{9}{16}W=\frac{9}{16}\times 48[W=48\text{ N}]$
$=27\text{ N}$