Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit of radius 3RE by supplying alpha x 106J of energy. The value of alpha is ____. (Take Radius of Earth RE = 6 x 106 m and g = 10 m/s2)

Energy of a satellite in a circular orbit is given as E = (-GME m/2r) where r is the radius of the circular orbit. Required energy to be supplied Delta E = Ef - Ei: Substitute GME = g RE2 into the expression: Comparing with alpha x 106 J, we get alpha = 1000. Orbital transition energy is always Delta E = (GMm/2) ((1/ri) - (1/rf)). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). Class 11 Physics: Gravitation

Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit of radius 3RE by supplying alpha x 106J of energy. The value of alpha is ____. (

Energy of a satellite in a circular orbit is given as E = (-GME m/2r) where r is the radius of the circular orbit. Required energy to be supplied Delta E = Ef - Ei: Substitute GME = g RE2 into the expression: Comparing with alpha x 106 J, we get alpha = 1000. Orbital transition energy is always Delta E = (GMm/2) ((1/ri) - (1/rf)). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). Class 11 Physics: Gravitation

Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit of radius 3RE by supplying alpha x 106J of energy. The value of alpha is ____. (Take Radius of Earth RE = 6 x 106 m and g = 10 m/s2)

Energy of a satellite in a circular orbit is given as E = (-GME m/2r) where r is the radius of the circular orbit. Required energy to be supplied Delta E = Ef - Ei: Substitute GME = g RE2 into the expression: Comparing with alpha x 106 J, we get alpha = 1000. Orbital transition energy is always Delta E = (GMm/2) ((1/ri) - (1/rf)). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). Class 11 Physics: Gravitation

Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit of radius 3RE by supplying alpha x 106J of energy. The value of alpha is ____. (

Energy of a satellite in a circular orbit is given as E = (-GME m/2r) where r is the radius of the circular orbit. Required energy to be supplied Delta E = Ef - Ei: Substitute GME = g RE2 into the expression: Comparing with alpha x 106 J, we get alpha = 1000. Orbital transition energy is always Delta E = (GMm/2) ((1/ri) - (1/rf)). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). Class 11 Physics: Gravitation

Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit…

Year	2026	2025	2024	Total
Questions	5	9	8	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

E = (-GME m)/(2r)

Δ E = Ef - Eᵢ

g = (GME)/(RE²) GME = g RE²

Core Logic

Energy of a satellite in a circular orbit is given as $E = (-GME m)/(2r)$ $E = \frac{-GM_E m}{2r}$ where r is the radius of the circular orbit. Required energy to be supplied $Δ E = Ef - Eᵢ$ $\Delta E = E_f - E_i$:

Δ E = ( (-GME m)/(2(3RE)) ) - ( (-GME m)/(2(1.5RE)) )

Δ E = (-GME m)/(6RE) + (GME m)/(3RE) = (GME m)/(6RE)

Step 1: Evaluate Delta E

Substitute $GME = g RE²$ $GM_E = g R_E^2$ into the expression:

Δ E = ((g RE²) m)/(6RE) = (1)/(6) m g RE

= (1)/(6) \times 100 \times 10 \times (6 \times 10⁶)

= 1000 \times 10⁶ J

Comparing with $α \times 10⁶ J$ $\alpha \times 10^6\text{ J}$, we get $α = 1000$ $\alpha = 1000$.

Pattern Recognition

Orbital transition energy is always $Δ E = (GMm)/(2) ((1)/(rᵢ) - (1)/(rf))$ . Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit).

Chapter Mix

Class 11 Physics: Gravitation

More Gravitation Previous-Year Questions — Page 5

Q46 jee_main_2024_31_jan_evening Escape Velocity

The mass of the moon is

1/144

times the mass of a planet and its diameter

1/16

times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

A. $(v)/(3)$ $\frac{v}{3}$
B. $(v)/(4)$ $\frac{v}{4}$
C. $(v)/(12)$ $\frac{v}{12}$
D. $(v)/(6)$ $\frac{v}{6}$

Solution

Related Formula

vescape = \sqrt((2GM)/(R))

Core Logic

For the planet: $v = \sqrt((2GMₚ)/(Rₚ))$ $v = \sqrt{\frac{2GM_p}{R_p}}$ For the moon: $Mm = (Mₚ)/(144)$ $M_m = \frac{M_p}{144}$ and $Rm = (Rₚ)/(16)$ $R_m = \frac{R_p}{16}$.

Step 1: Setup the Ratio

vm = \sqrt((2G Mm)/(Rm))

vm = \sqrt((2G ((Mₚ)/(144)))/(((Rₚ)/(16))))

vm = \sqrt((2G Mₚ)/(Rₚ) \times (16)/(144))

Step 2: Simplification

vm = \sqrt((2G Mₚ)/(Rₚ)) \times \sqrt((1)/(9))

vm = v \times (1)/(3) = (v)/(3)

Pattern Recognition

Escape velocity scales as $\sqrt(M/R)$ $\sqrt{M/R}$. If $M$ $M$ scales by $x$ $x$ and $R$ $R$ scales by $y$ $y$, velocity scales by $\sqrt(x/y)$ $\sqrt{x/y}$. Here, $\sqrt((1/144)/(1/16)) = \sqrt(16/144) = \sqrt(1/9) = 1/3$ .

Chapter Mix

Class 11 Physics: Gravitation

Q jee_main_2024_31_jan_morning Superposition Principle

Four identical particles of mass

m

are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is

( 2\sqrt(2) + 132) Gm²L²

, the length of the sides of the square is

A. $L2$ $\frac{\mathrm{L}}{2}$
B. $4 L$ $4 L$
C. $3L$ $3L$
D. $2 L$ $2 L$

Solution

Related Formula

F = (G m₁ m₂)/(r²)

Core Logic

Superposition Principle diagram for Q37 - JEE Main 2024 Morning

Let the side length of the square be $a$ $a$. Considering one corner mass, it experiences forces from the adjacent two masses (distance $a$ $a$) and the diagonally opposite mass (distance $\sqrt(2)a$ $\sqrt{2}a$).

The forces from the two adjacent masses are at $90^°$ $90^\circ$ to each other:

F = (Gm²)/(a²)

The resultant of these two is $\sqrt(2)F = \sqrt(2) (Gm²)/(a²)$ $\sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$, directed along the diagonal.

Step 2: Total Force Equation

The force from the diagonal mass is:

F' = Gm²(\sqrt(2)a)² = (Gm²)/(2a²)

Total resultant force $Fₙₑₜ = \sqrt(2)F + F'$ $F_{\text{net}} = \sqrt{2}F + F'$:

Fₙₑₜ = \sqrt(2) (Gm²)/(a²) + (Gm²)/(2a²) = (Gm²)/(a²) ( \sqrt(2) + (1)/(2) )

Fₙₑₜ = (Gm²)/(a²) ( 2\sqrt(2) + 12 )

Equating this to the given force value:

( 2\sqrt(2) + 132)(Gm²)/(L²) = (Gm²)/(a²) ( 2\sqrt(2) + 12 )

(1)/(32 L²) = (1)/(2 a²)

$a² = 16 L²$ $a^2 = 16 L^2$ $a = 4L$ $a = 4L$

Chapter Mix

Class 11 Physics: Gravitation

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Delta E

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Setup the Ratio

Step 2: Simplification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Total Force Equation

Chapter Mix

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Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Delta E

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Setup the Ratio

Step 2: Simplification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Total Force Equation

Chapter Mix

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