Q.Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A 1200 kg artificial satellite is in an orbit of radius $2R_E$ about the earth. The energy required to transfer it to an orbit of radius $3R_E$ is ($g = 10\ ms^{-2}$, $R_E = 6400\ km$) (A) $1.2 \times 10^9\ J$ (B) $6.4 \times 10^9\ J$ (C) $6400 \times 10^3\ J$ (D) $3.2 \times 10^9\ J$

   ›Reveal solutionSolution

   Tests total orbital energy and the energy needed to move a satellite to a higher orbit. Answer: $6.4\times10^9$ J.

   Concept and Intuition

   A satellite in a circular orbit has total mechanical energy $E=-\dfrac{GMm}{2r}$ (negative, bound orbit), which becomes less negative (i.e. increases) as $r$ increases — moving to a higher orbit requires adding energy, even though the satellite ends up moving slower there, because potential energy increases faster than kinetic energy decreases.

   Step-by-Step Solution

   1. Energy in orbit of radius $r$: $E(r)=-\dfrac{GMm}{2r}$.
   2. Required energy: $\Delta E = E(3R_E)-E(2R_E) = -\dfrac{GMm}{6R_E}+\dfrac{GMm}{4R_E}=GMm\left(\dfrac14-\dfrac16\right)\dfrac{1}{R_E}=\dfrac{GMm}{12R_E}$.
   3. Use $GM=gR_E^2$ (from $g=GM/R_E^2$ at the surface): $GM=10\times(6.4\times10^6)^2=10\times4.096\times10^{13}=4.096\times10^{14}\ m^3s^{-2}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Work required to shift an artificial satellite from an orbit of radius '$r$' to an orbit of radius '$2r$' is (A) $\dfrac{GMm}{2r}$ (B) $\dfrac{GMm}{4r}$ (C) $\dfrac{GMm}{8r}$ (D) zero

   ›Reveal solutionSolution

   This tests the total energy formula for a satellite in circular orbit, and that the work needed to change orbits equals the difference in total orbital energies. The answer is $\dfrac{GMm}{4r}$.

   Concept and Intuition

   For a satellite of mass $m$ orbiting a planet of mass $M$ in a circular orbit of radius $r$, the total mechanical energy (kinetic + potential) is $E = -\dfrac{GMm}{2r}$ — negative because the satellite is gravitationally bound. Moving to a larger orbit radius means a less negative (higher) total energy, so external work must be done on the satellite. This work equals the difference between the final and initial total energies (by the work-energy theorem applied to the whole orbit-raising process).

   Step-by-Step Solution

   1. Total energy in orbit of radius $r$: $E_i = -\dfrac{GMm}{2r}$.
   2. Total energy in orbit of radius $2r$: $E_f = -\dfrac{GMm}{2(2r)} = -\dfrac{GMm}{4r}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If a body is thrown vertically upwards from the surface of the earth with a speed equal to 75% of the escape speed from the surface of the earth, then the ratio of the maximum height reached by the body and the radius of the earth is (A) 5 : 7 (B) 9 : 7 (C) 3 : 7 (D) 11 : 7

   ›Reveal solutionSolution

   Because the launch speed (75% of escape speed) is a large fraction of $v_e$, the height reached is comparable to Earth's radius, so we must use the exact (varying-$g$) gravitational PE, not $mgh$. Energy conservation gives $h/R = 9/7$.

   Concept and Intuition

   The familiar $H = u^2/2g$ formula assumes gravity is constant over the rise — valid only for heights much smaller than the Earth's radius. Here the speed is a substantial fraction of the escape speed, so the body could rise to a height comparable to $R$ itself, where $g$ noticeably weakens. We must instead use the full gravitational potential energy $U(r) = -GM/r$ and conserve total mechanical energy between the surface and the highest point (where the radial velocity is momentarily zero).

   Step-by-Step Solution

   1. Escape speed: $v_e^2 = \dfrac{2GM}{R}$.
   2. Launch speed: $v = 0.75\,v_e \Rightarrow v^2 = 0.5625\,v_e^2 = 0.5625\times\dfrac{2GM}{R} = 1.125\dfrac{GM}{R}$.
   3. Energy conservation from the surface (radius $R$, speed $v$) to the highest point (radius $R+h$, speed $0$): $\tfrac12 v^2 - \dfrac{GM}{R} = 0 - \dfrac{GM}{R+h}$ …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A rocket is fired vertically from the surface of the earth with quarter the escape speed. If R is radius of the earth, the maximum altitude reached by the rocket is (A) $\dfrac{R}{5}$ (B) $\dfrac{R}{3}$ (C) $\dfrac{R}{15}$ (D) $\dfrac{R}{14}$

   ›Reveal solutionSolution

   Tests energy conservation for a projectile launched at a fraction of escape speed to find the maximum height reached (using the exact $1/r$ potential, not $mgh$).

   Concept and Intuition

   Escape speed is the speed needed to just reach infinity with zero leftover kinetic energy. A rocket launched slower than escape speed will rise, decelerate under gravity, and momentarily stop at some finite maximum altitude — found by conserving total mechanical energy between the surface and that highest point (where KE=0). Because the altitude is not small compared to $R$, we must use the exact gravitational PE $-GM m/r$, not the uniform-field approximation $mgh$.

   Step-by-Step Solution

   1. Escape speed: $v_e = \sqrt{\dfrac{2GM}{R}}$. Launch speed $v = \dfrac{v_e}{4}$, so $v^2 = \dfrac{v_e^2}{16} = \dfrac{2GM}{16R} = \dfrac{GM}{8R}$.
   2. Energy conservation from surface (radius $R$) to maximum altitude (radius $R+h$, velocity $=0$):

   $\dfrac12 m v^2 - \dfrac{GMm}{R} = 0 - \dfrac{GMm}{R+h}$

   3. Divide by $m$ and substitute $v^2$:

   $\dfrac{GM}{16R} - \dfrac{GM}{R} = -\dfrac{GM}{R+h}$

   4. Simplify the left side: $\dfrac{GM}{16R} - \dfrac{16GM}{16R} = -\dfrac{15GM}{16R}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.A satellite of mass 1000 kg is revolving around the earth at height equal to $\dfrac{R}{4}$. Then the energy to be given to the satellite to revolve around the earth at height $\dfrac{R}{2}$ is (Acceleration due to gravity 10 m s$^{-2}$, Radius of the earth R=6400 km) (A) $64 \times 10^9$ J (B) $32 \times 10^9$ J (C) $96 \times 10^9$ J (D) $16 \times 10^9$ J

   ›Reveal solutionSolution

   The energy needed to move a satellite between two orbits is the difference of total orbital energies $E=-GMm/2r$; plugging in the given radii and $GM=gR^2$ gives $64\times10^9$ J. Answer: (A).

   Concept and Intuition

   A satellite in a bound circular orbit has total mechanical energy $E=-\dfrac{GMm}{2r}$ (negative, since it's bound). Moving it to a larger orbital radius requires supplying energy — even though its kinetic energy actually decreases at the larger radius (it orbits slower), the potential energy increases by more than enough to cover both the KE decrease and the net energy that must be added. The clean way to find "energy to be given" is simply the difference of the total energies at the two radii.

   Step-by-Step Solution

   1. Total orbital energy at radius $r$: $E(r)=-\dfrac{GMm}{2r}$.
   2. Energy required $\Delta E = E(r_2)-E(r_1) = \dfrac{GMm}{2}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$.
   3. With $r_1=R/4$ and $r_2=R/2$: $\dfrac1{r_1}=\dfrac4R$, $\dfrac1{r_2}=\dfrac2R$, difference $=\dfrac2R$.
   4. $\Delta E = \dfrac{GMm}{2}\times\dfrac{2}{R}=\dfrac{GMm}{R}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.An infinite number of objects each 1 kg mass are placed on the x-axis on both sides of x = 0 at $\pm1\ \mathrm{m}, \pm2\ \mathrm{m}, \pm4\ \mathrm{m}, \pm8\ \mathrm{m} \ldots$ and so on. The magnitude of the resultant gravitational potential (in SI units) at x = 0 is (G - Universal gravitational constant) (A) $-G$ (B) $-2G$ (C) $-3G$ (D) $-4G$

   ›Reveal solutionSolution

   Superposition of gravitational potential from an infinite geometric arrangement of point masses; the geometric series sums to give $V=-4G$.

   Concept and Intuition

   Gravitational potential is a scalar, so contributions from every mass simply add (with sign). Potential due to a point mass $m$ at distance $r$ is $V=-\dfrac{Gm}{r}$. Here masses of 1 kg are placed at distances $1,2,4,8,\ldots$ m on both sides of the origin — a geometric progression with common ratio $\tfrac12$ (in terms of $1/r$), which sums to a finite value even though there are infinitely many masses.

   Step-by-Step Solution

   1. On the positive $x$-axis, masses of 1 kg sit at $r=1,2,4,8,\ldots = 2^n$ m for $n=0,1,2,\ldots$.
   2. Contribution from positive side: $-G\sum_{n=0}^{\infty}\dfrac{1}{2^n} = -G\left(\dfrac{1}{1-\tfrac12}\right) = -2G$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A mass of $6 \times 10^{24}$ kg is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4\ \mathrm{m\,s^{-1}}$. The radius of the sphere is (Universal gravitational constant $= 6.66 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$) (A) 483 km (B) 575 km (C) 789 km (D) 888 km

   ›Reveal solutionSolution

   Tests the escape-velocity formula $v_e=\sqrt{2GM/R}$ rearranged to find the radius a given mass must be compressed to. Answer: 888 km.

   Concept and Intuition

   Escape velocity is the minimum speed needed for an object to leave a gravitating body's surface without further propulsion, found by equating kinetic energy to the magnitude of gravitational potential energy: $\tfrac{1}{2}mv_e^2=\dfrac{GMm}{R}$, giving $v_e=\sqrt{\dfrac{2GM}{R}}$. If a fixed mass is compressed into a smaller sphere, $R$ decreases and $v_e$ increases — this is the physical idea behind black holes (compress enough mass into a small enough radius and even light cannot escape).

   Step-by-Step Solution

   1. Start from $v_e=\sqrt{\dfrac{2GM}{R}}$.
   2. Square both sides: $v_e^2=\dfrac{2GM}{R}$.
   3. Solve for $R$: $R=\dfrac{2GM}{v_e^2}$.
   4. Substitute values: $2GM = 2\times6.66\times10^{-11}\times6\times10^{24}=7.992\times10^{14}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.If an object of mass 1 kg is taken to a height which is equal to three times the radius of the earth, then the change in its potential energy is (Radius of the earth = 6400 km, acceleration due to gravity on the surface of the earth = 10 m s$^{-2}$) (A) $48 \times 10^6$ J (B) $24 \times 10^6$ J (C) $36 \times 10^6$ J (D) $12 \times 10^6$ J

   ›Reveal solutionSolution

   Since the height (3R) is comparable to Earth's radius, the exact gravitational PE formula $-GMm/r$ must be used instead of $mgh$; the change works out to $48\times10^6$ J.

   Concept and Intuition

   $mgh$ only applies near the surface where $g$ is essentially constant. Here the object is lifted by $3R$ — far too large a distance for that approximation — so we must use the exact potential energy $U(r) = -\dfrac{GMm}{r}$, measured from the center of the Earth.

   Step-by-Step Solution

   1. Initial distance from Earth's center: $R$. Final distance: $R+3R=4R$.
   2. $\Delta PE = U(4R)-U(R) = -\dfrac{GMm}{4R}-\left(-\dfrac{GMm}{R}\right) = GMm\left(\dfrac1R-\dfrac1{4R}\right) = \dfrac{3GMm}{4R}$.
   3. Using $g=\dfrac{GM}{R^2}$, we get $\dfrac{GM}{R}=gR$, so $\Delta PE = \dfrac{3mgR}{4}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.If a body is projected from the surface of the earth with a velocity of $\sqrt{5}\,V_E$, then the velocity of the body when it escapes from the gravitational influence of the earth is (Escape velocity of a body from the surface of the earth, $V_E = 11.2\ \mathrm{km\,s^{-1}}$) (A) $22.4\ \mathrm{km\,s^{-1}}$ (B) $11.2\ \mathrm{km\,s^{-1}}$ (C) $11.2\sqrt{5}\ \mathrm{km\,s^{-1}}$ (D) $5.6\ \mathrm{km\,s^{-1}}$

   ›Reveal solutionSolution

   Applying energy conservation with the definition of escape velocity, the residual speed at infinity works out to exactly $2V_E = 22.4$ km/s.

   Concept and Intuition

   Escape velocity $V_E$ is defined as the minimum launch speed for which a body just barely escapes Earth's gravity, arriving at infinity with zero residual speed: $\frac12 V_E^2 = \frac{GM}{R}$ (all its kinetic energy is used up climbing out of the potential well). If a body is launched faster than $V_E$, it will still have some leftover kinetic energy (hence leftover speed) once it's escaped to infinity, found by simply subtracting the potential-well "toll" ($\frac{GM}{R} = \frac12V_E^2$) from its initial kinetic energy per unit mass.

   Step-by-Step Solution

   1. Energy conservation per unit mass (ignoring rotation, launched radially): $\dfrac12 v_0^2 - \dfrac{GM}{R} = \dfrac12 v_\infty^2$.
   2. Substitute $\dfrac{GM}{R} = \dfrac12 V_E^2$: $\dfrac12 v_0^2 - \dfrac12 V_E^2 = \dfrac12 v_\infty^2$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.If the escape velocity of a body from the surface of the earth is $11.2 \mathrm{~km~s^{-1}}$, then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the earth is (A) $11.2 \mathrm{~km~s^{-1}}$ (B) $2.8 \mathrm{~km~s^{-1}}$ (C) $22.4 \mathrm{~km~s^{-1}}$ (D) $5.6 \mathrm{~km~s^{-1}}$

    ›Reveal solutionSolution

    This tests the relationship between escape velocity and orbital velocity at a given altitude; the answer is (D) $5.6\ \mathrm{km\,s^{-1}}$.

    Concept and Intuition

    Escape velocity from the surface is $v_e=\sqrt{2gR}$, derived from energy conservation. Orbital velocity at height $h$ (orbit radius $R+h$) comes from equating gravitational force to the centripetal requirement: $v_o = \sqrt{\dfrac{gR^2}{R+h}}$. Both expressions share the combination $gR$, which lets us eliminate $g$ and $R$ separately and express $v_o$ purely in terms of $v_e$.

    Step-by-Step Solution

    1. $v_e = \sqrt{2gR} \Rightarrow gR = \dfrac{v_e^2}{2}$.
    2. At height $h=R$, orbit radius is $2R$: $v_o = \sqrt{\dfrac{gR^2}{2R}} = \sqrt{\dfrac{gR}{2}}$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The potential energy of a satellite of mass 'm' revolving around the earth at a height of $R_e$ from the surface of the earth is ($R_e$ - radius of earth; g - acceleration due to gravity) (A) $-0.5\,mgR_e$ (B) $-mgR_e$ (C) $-2mgR_e$ (D) $-4mgR_e$

    ›Reveal solutionSolution

    Substituting $GM=gR_e^2$ into the gravitational PE formula at orbital radius $2R_e$ gives $-0.5\,mgR_e$.

    Concept and Intuition

    Gravitational potential energy of a satellite at distance $r$ from Earth's centre is $U=-\dfrac{GMm}{r}$. Since $g=\dfrac{GM}{R_e^2}$ at the surface, we can re-express $GM$ in terms of $g$ and $R_e$ to avoid needing $G$ and $M$ separately.

    Step-by-Step Solution

    1. Height above surface $=R_e$, so distance from Earth's centre $r = R_e + R_e = 2R_e$.
    2. Gravitational PE: $U = -\dfrac{GMm}{r} = -\dfrac{GMm}{2R_e}$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.The escape velocity of a body from a planet of mass M and radius R is $14\ \mathrm{km\,s^{-1}}$. The escape velocity of the body from another planet having same mass and diameter 8R (in $\mathrm{km\,s^{-1}}$) is (A) 7 (B) 10.5 (C) 14 (D) 28

    ›Reveal solutionSolution

    With the same mass but 4 times the radius, escape velocity (which scales as $1/\sqrt{R}$) is halved: 7 km/s.

    Concept and Intuition

    Escape velocity depends on both the mass and the radius of the body being escaped from: $v_{esc}=\sqrt{\dfrac{2GM}{R}}$. Same mass but a larger radius means the surface is farther from the centre, so gravity there is weaker and less speed is needed to escape.

    Step-by-Step Solution

    1. Given $v_{esc}=\sqrt{2GM/R}=14\,\text{km/s}$ for radius $R$.
    2. New planet: same mass $M$, diameter $8R$ ⇒ radius $=4R$. …

    🔒 Start your 14-day free trial to see the full solution →