Gravitational Field and Orbits Practice Questions — JAMB UTME

Question 1Question →

Two point masses are separated by a distance $r$ and exert a gravitational force $F$ on each other. If the distance between them is doubled while keeping their masses constant, what is the new gravitational force between them?

\frac{F}{4}

\frac{F}{2}

2F

4F

Show answer & explanation

Answer: $\frac{F}{4}$

Answer

The new gravitational force between the two masses is

\frac{F}{4}

.

Gravitational force obeys an inverse-square law with respect to distance (

F \propto \frac{1}{r^2}

). When separation distance is multiplied by 2, the force decreases by a factor of

2^2 = 4

, yielding

\frac{F}{4}

.

Step-by-Step Solution

1

Write down Newton's Law of Universal Gravitation

F = \frac{G m_1 m_2}{r^2}

Establish the mathematical relationship governing gravitational force and separation distance.

2

Substitute the new distance

r' = 2r

into the gravitational force equation

F' = \frac{G m_1 m_2}{(2r)^2} = \frac{G m_1 m_2}{4r^2}

Evaluate how doubling the separation distance affects the magnitude of the force.

3

Express the new force

F'

in terms of the initial force

F

F' = \frac{1}{4}\left(\frac{G m_1 m_2}{r^2}\right) = \frac{F}{4}

Relate the calculated force directly to the original force

F

.

Key Concept

Inverse-Square Law of Gravitation

Estimated Time:45s

Question 2Question →

A satellite of mass $m$ moves in a circular orbit around a uniform spherical planet of radius $R$ . If the height of the satellite above the planet's surface is $h = 2R$ and the acceleration due to gravity at the planet's surface is $g$ , which of the following expressions represents the orbital speed of the satellite?

\sqrt{\frac{gR}{3}}

\sqrt{\frac{gR}{2}}

\sqrt{\frac{2gR}{3}}

\sqrt{3gR}

Show answer & explanation

Answer: $\sqrt{\frac{gR}{3}}$

Answer

The orbital speed of the satellite is

\sqrt{\frac{gR}{3}}

.

The total orbital radius from the center of the planet is

r = R + h = R + 2R = 3R

. Since surface gravity is

g = \frac{GM}{R^2}

, we have

GM = gR^2

. Substituting these into the orbital velocity expression

v = \sqrt{\frac{GM}{r}}

yields

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

, which makes the expression

\sqrt{\frac{gR}{3}}

correct.

Step-by-Step Solution

1

Determine the total orbital radius from the center of the planet.

r = R + h = R + 2R = 3R

Gravitational attraction and circular orbital radii are always measured from the center of mass of the primary body, not its surface.

2

Relate the gravitational constant

G

and planet mass

M

to surface gravity

g

.

g = \frac{GM}{R^2} \implies GM = gR^2

At the surface of a spherical planet of radius

R

, the gravitational field strength is

g

.

3

Substitute

r = 3R

and

GM = gR^2

into the circular orbital velocity equation

v = \sqrt{\frac{GM}{r}}

.

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

Equating centripetal force to gravitational force

\frac{m v^2}{r} = \frac{G M m}{r^2}

yields

v = \sqrt{\frac{GM}{r}}

.

Key Concept

Orbital velocity of a satellite in terms of surface gravitational acceleration and orbital radius

Question 3Question →

A uniform spherical planet has twice the mass and twice the radius of the Earth. If the acceleration due to gravity at the Earth's surface is $g$ , what is the acceleration due to gravity at the surface of this planet?

0.25g

0.5g

1.0g

2.0g

Show answer & explanation

Answer: $0.5g$

Answer

The acceleration due to gravity at the surface of the planet is

0.5g

.

The acceleration due to gravity at a planet's surface is given by

g = \frac{GM}{R^2}

. Doubling the mass doubles the field strength, but doubling the radius reduces the field strength by a factor of

2^2 = 4

due to the inverse-square law. Combining these changes results in

\frac{2}{4}g = 0.5g

.

Step-by-Step Solution

1

Write the general expression for surface acceleration due to gravity.

g = \frac{GM}{R^2}

, where

G

is the gravitational constant,

M

is planetary mass, and

R

is planetary radius.

Establishes the fundamental formula relating gravity to mass and radius.

2

Substitute the given parameters for the new planet (

M_p = 2M

and

R_p = 2R

).

g_p = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2}

.

Applies the proportional changes to mass and radius into the formula.

3

Simplify the expression to find

g_p

in terms of

g

.

g_p = \frac{2}{4} \left(\frac{GM}{R^2}\right) = 0.5g

.

Evaluates the fractional change relative to Earth's surface gravity.

Key Concept

Dependence of surface gravitational field strength on planetary mass and radius via the inverse-square law.

Estimated Time:1m 15s

Question 4Question →

A spherical planet has a radius of $7.2 \times 10^{6} \text{ m}$ and an acceleration due to gravity of $10 \text{ m/s}^2$ at its surface. What is the escape velocity for an object launched from the surface of this planet, in $\text{km/s}$ ?

Show answer & explanation

Answer: 12

Answer

The escape velocity from the surface of the planet is 12 km/s.

The escape velocity

v_e

from the surface of a spherical planet of radius

R

with surface gravity

g

is given by

v_e = \sqrt{2gR}

. Substituting

g = 10 \text{ m/s}^2

and

R = 7.2 \times 10^6 \text{ m}

yields

v_e = \sqrt{2 \times 10 \times 7.2 \times 10^6} = \sqrt{1.44 \times 10^8} = 12000 \text{ m/s}

, which equals

12 \text{ km/s}

.

Step-by-Step Solution

1

Identify the relationship between surface gravity, planetary radius, and escape velocity

The escape velocity formula is

v_e = \sqrt{2gR}

.

Escape velocity is the minimum initial speed required for an object to overcome the gravitational pull of a celestial body.

2

Substitute the given numerical values into the formula

v_e = \sqrt{2 \times 10 \text{ m/s}^2 \times 7.2 \times 10^6 \text{ m}} = \sqrt{144 \times 10^6} \text{ m/s}

.

Plugging in the given values allows direct calculation of the velocity in standard SI units.

3

Simplify the square root and convert units to km/s

v_e = 12000 \text{ m/s} = 12 \text{ km/s}

.

Taking the square root of

144 \times 10^6

gives

12000 \text{ m/s}

, which corresponds to

12 \text{ km/s}

.

Key Concept

Escape Velocity from a Planet's Surface

Question 5Question →

A satellite orbits the Earth at an altitude equal to three times the radius of the Earth, $R$ . If the acceleration due to gravity at the Earth's surface is $g$ , what is the gravitational field strength experienced by the satellite at its orbit?

\frac{g}{16}

\frac{g}{9}

\frac{g}{4}

\frac{g}{3}

Show answer & explanation

Answer: $\frac{g}{16}$

Answer

The gravitational field strength at the orbital position is

\frac{g}{16}

.

The total distance from the center of the Earth to the satellite is

r = R + 3R = 4R

. Since gravitational field strength is inversely proportional to the square of the distance from the center of mass, increasing the distance by a factor of

4

reduces the field strength by a factor of

4^2 = 16

, giving

\frac{g}{16}

.

Step-by-Step Solution

1

Determine the total distance from the center of the Earth to the satellite.

The orbital radius is

r = R + h = R + 3R = 4R

.

Gravitational calculations must be measured from the center of mass of the attracting body, not its surface.

2

Apply Newton's law of universal gravitation for gravitational field strength.

At the surface (

r = R

),

g = \frac{GM}{R^2}

. At orbit (

r = 4R

),

g' = \frac{GM}{(4R)^2} = \frac{GM}{16R^2}

.

Gravitational field strength follows an inverse-square law with respect to distance from the center of mass.

3

Express the orbital gravitational field strength in terms of the surface gravity

g

.

g' = \frac{1}{16} \left(\frac{GM}{R^2}\right) = \frac{g}{16}

.

Substituting the surface value

g = \frac{GM}{R^2}

yields the simplified ratio.

Key Concept

Gravitational Field Strength and Inverse-Square Law

Question 6Question →

Planet $P$ has a mean uniform density that is twice that of Planet $Q$ , and a radius that is half that of Planet $Q$ . What is the ratio of the escape velocity at the surface of Planet $P$ to the escape velocity at the surface of Planet $Q$ ?

\frac{1}{\sqrt{2}}

\frac{1}{2}

1

\sqrt{2}

Show answer & explanation

Answer: $\frac{1}{\sqrt{2}}$

Answer

\frac{1}{\sqrt{2}}

The correct answer is derived by substituting the planet's mass in terms of radius and density (

M = \frac{4}{3}\pi R^3 \rho

) into the escape velocity equation

v_e = \sqrt{\frac{2GM}{R}}

. Simplifying gives

v_e = R\sqrt{\frac{8}{3}\pi G \rho}

, which means escape velocity is proportional to

R\sqrt{\rho}

. Halving the radius and doubling the density yields a factor of

\frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

Step-by-Step Solution

1

Express the mass of a uniform spherical planet in terms of its radius

R

and density

\rho

.

M = \frac{4}{3}\pi R^3 \rho

Mass equals volume multiplied by mean density.

2

Substitute mass into the escape velocity formula

v_e = \sqrt{\frac{2GM}{R}}

.

v_e = \sqrt{\frac{2G \left(\frac{4}{3}\pi R^3 \rho\right)}{R}} = R\sqrt{\frac{8}{3}\pi G \rho}

This establishes the proportionality

v_e \propto R\sqrt{\rho}

.

3

Set up the ratio of escape velocity for Planet

P

to Planet

Q

using

R_P = \frac{1}{2}R_Q

and

\rho_P = 2\rho_Q

.

\frac{v_{e,P}}{v_{e,Q}} = \frac{R_P\sqrt{\rho_P}}{R_Q\sqrt{\rho_Q}} = \frac{\left(\frac{1}{2}R_Q\right)\sqrt{2\rho_Q}}{R_Q\sqrt{\rho_Q}} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}}

Evaluating the ratio of proportional quantities yields the final scale factor.

Key Concept

Dependence of Escape Velocity on Planet Density and Radius

Estimated Time:2m 0s

Question 7Question →

A satellite revolves around a planet in a circular orbit of radius $1.0 \times 10^4 \text{ km}$ with an orbital period of $12 \text{ hours}$ . Calculate the orbital period, in hours, of a second satellite orbiting the same planet in a circular path of radius $4.0 \times 10^4 \text{ km}$ .

Show answer & explanation

Answer: 96

Answer

96 hours

According to Kepler's Third Law (

T^2 \propto r^3

), the orbital period

T

scales with radius

r

as

T \propto r^{3/2}

. Increasing the orbital radius by a factor of 4 increases the period by a factor of

4^{3/2} = 8

. Multiplying the original period of 12 hours by 8 yields 96 hours.

Step-by-Step Solution

1

Set up Kepler's Third Law equation relating orbital period and orbital radius.

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

Kepler's Third Law states that the square of the orbital period of a body in circular orbit is directly proportional to the cube of the radius of its orbit.

2

Substitute the given orbital radii and evaluate the scaling factor.

\frac{r_2}{r_1} = \frac{4.0 \times 10^4 \text{ km}}{1.0 \times 10^4 \text{ km}} = 4

Simplifying the ratio of the two orbital radii gives a factor of 4 increase in radius.

3

Calculate the period multiplier by taking the ratio to the power of 3/2.

4^{3/2} = (\sqrt{4})^3 = 2^3 = 8

Taking

T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2}

shows the period scales by a factor of 8.

4

Multiply the initial orbital period by the scaling factor to find the final answer.

T_2 = 12 \text{ hours} \times 8 = 96 \text{ hours}

Multiplying the baseline period of 12 hours by 8 yields the new orbital period.

Key Concept

Kepler's Third Law of Planetary Motion

Estimated Time:1m 30s

Question 8Question →

A satellite of mass $500\text{ kg}$ orbits a spherical planet of radius $R = 6.0 \times 10^6\text{ m}$ with surface gravitational acceleration $g = 10\text{ m/s}^2$ . The satellite is transferred from an initial circular orbit of radius $2R$ to a higher circular orbit of radius $3R$ . What is the minimum energy required, in megajoules ( $\text{MJ}$ ), to perform this transfer?

Show answer & explanation

Answer: 2500

Answer

The minimum energy required to perform the orbital transfer is

2500\text{ MJ}

.

The minimum energy needed to move a satellite between circular orbits is equal to the change in its total mechanical energy (

E = -\frac{GMm}{2r}

). Expressing

GM

as

gR^2

, the energy difference between radii

2R

and

3R

simplifies to

\Delta E = \frac{gRm}{12}

, which evaluates to

2500\text{ MJ}

.

Step-by-Step Solution

1

Relate surface acceleration due to gravity to planet mass and radius.

GM = gR^2

At the planet's surface (

r = R

), gravitational acceleration is

g = \frac{GM}{R^2}

.

2

Formulate the total mechanical energy equation for a circular orbit.

E = -\frac{GMm}{2r} = -\frac{gR^2 m}{2r}

Total energy is kinetic energy

\frac{GMm}{2r}

plus gravitational potential energy

-\frac{GMm}{r}

.

3

Calculate initial and final total energies.

E_1 = -\frac{gRm}{4}

and

E_2 = -\frac{gRm}{6}

Substitute the orbit radii

r_1 = 2R

and

r_2 = 3R

into the total energy equation.

4

Determine the net work required for the transfer.

\Delta E = E_2 - E_1 = \frac{gRm}{12}

The energy required equals the difference in total mechanical energy between the final and initial orbits.

5

Substitute given numerical values and convert joules to megajoules.

\Delta E = \frac{10 \times (6.0 \times 10^6) \times 500}{12} = 2.5 \times 10^9\text{ J} = 2500\text{ MJ}

Dividing

2.5 \times 10^9\text{ J}

by

10^6

converts the value to megajoules.

Key Concept

Total Mechanical Energy of a Satellite in Circular Orbit and Orbital Transfer Energy

Estimated Time:3m 0s

Question 9Question →

An astronaut has a weight of $720\text{ N}$ on the surface of the Earth. Calculate the weight of the astronaut, in Newtons ( $\text{N}$ ), at an altitude equal to twice the radius of the Earth ( $h = 2R$ ).

Show answer & explanation

Answer: 80

Answer

The weight of the astronaut at an altitude of

2R

is

80\text{ N}

.

At an altitude of

2R

, the total distance from the center of the Earth is

r = R + 2R = 3R

. Because gravitational force follows the inverse-square law (

W \propto 1/r^2

), tripling the distance reduces the gravitational force and weight by a factor of

3^2 = 9

. Dividing the surface weight of

720\text{ N}

by

9

yields

80\text{ N}

.

Step-by-Step Solution

1

Determine the total distance from the center of the Earth

r = R + 2R = 3R

Gravitational force depends on the distance measured from the center of mass of the Earth, which is the sum of Earth's radius

R

and altitude

h

.

2

Apply the inverse-square law of gravitation to find field strength at altitude

g' = \frac{g}{3^2} = \frac{g}{9}

Acceleration due to gravity is inversely proportional to the square of the distance from the planet's center (

g \propto \frac{1}{r^2}

).

3

Calculate the astronaut's weight at altitude

W' = \frac{720}{9} = 80\text{ N}

Weight is directly proportional to gravitational field strength (

W = mg

).

Key Concept

Variation of Acceleration due to Gravity with Altitude (Inverse Square Law)

Question 10Question →

Two point masses of $4.0\text{ kg}$ and $9.0\text{ kg}$ are separated by a distance of $5.0\text{ m}$ in free space. At what distance from the $4.0\text{ kg}$ mass along the line joining them is the net gravitational field strength equal to zero?

2.0\text{ m}

3.0\text{ m}

2.5\text{ m}

1.0\text{ m}

Show answer & explanation

Answer: $2.0\text{ m}$

Answer

The distance from the

4.0\text{ kg}

mass where the net gravitational field strength is zero is

2.0\text{ m}

.

At the point where the net gravitational field strength is zero, the gravitational field intensity produced by the

4.0\text{ kg}

mass must equal the intensity produced by the

9.0\text{ kg}

mass in magnitude. Equating

\frac{G (4.0)}{x^2} = \frac{G (9.0)}{(5.0 - x)^2}

and taking square roots yields

\frac{2}{x} = \frac{3}{5 - x}

. Cross-multiplying gives

10 - 2x = 3x

, which yields

x = 2.0\text{ m}

from the

4.0\text{ kg}

mass.

Step-by-Step Solution

1

Set up the condition for zero net gravitational field strength.

E_1 = E_2 \implies \frac{G m_1}{x^2} = \frac{G m_2}{(d - x)^2}

At the neutral point, the opposing gravitational field vectors due to both masses are equal in magnitude.

2

Substitute given values

m_1 = 4.0\text{ kg}

,

m_2 = 9.0\text{ kg}

, and total distance

d = 5.0\text{ m}

.

\frac{4.0}{x^2} = \frac{9.0}{(5.0 - x)^2}

Simplifying by canceling

G

from both sides of the equation.

3

Take the square root of both sides and solve for

x

.

\frac{2.0}{x} = \frac{3.0}{5.0 - x} \implies 2.0(5.0 - x) = 3.0x \implies 10.0 - 2.0x = 3.0x \implies 5.0x = 10.0 \implies x = 2.0\text{ m}

Taking the square root simplifies the quadratic relationship into a linear ratio.

Key Concept

Gravitational Field Strength Neutral Point

Estimated Time:1m 0s

Question 11Question →

Two satellites, $X$ and $Y$ , move in circular orbits around a planet with orbital radii of $r$ and $4r$ , respectively. If satellite $X$ travels at an orbital speed of $v$ , what is the orbital speed of satellite $Y$ ?

\frac{v}{2}

\frac{v}{4}

2v

4v

Show answer & explanation

Answer: $\frac{v}{2}$

Answer

The orbital speed of satellite

Y

is

\frac{v}{2}

.

The orbital speed

v

of a satellite in a circular orbit of radius

r

around a planet of mass

M

is given by

v = \sqrt{\frac{GM}{r}}

. When the radius increases by a factor of 4 from

r

to

4r

, the new speed becomes

v_Y = \sqrt{\frac{GM}{4r}} = \frac{1}{2} \sqrt{\frac{GM}{r}} = \frac{v}{2}

.

Step-by-Step Solution

1

Write the formula for orbital speed of a satellite

v = \sqrt{\frac{GM}{r}}

Orbital speed depends on the gravitational mass

M

of the central body and the orbital radius

r

.

2

Set up the ratio between the orbital speeds of satellites

Y

and

X

\frac{v_Y}{v_X} = \sqrt{\frac{r_X}{r_Y}}

Since

G

and

M

are constant, orbital speed is inversely proportional to

\sqrt{r}

.

3

Substitute the given values

r_X = r

,

r_Y = 4r

, and

v_X = v

v_Y = v \sqrt{\frac{r}{4r}} = v \sqrt{\frac{1}{4}} = \frac{v}{2}

Evaluating the square root yields a factor of

\frac{1}{2}

.

Key Concept

Orbital Speed of a Satellite

Estimated Time:1m 0s

Question 12Question →

An object of mass $m$ is projected vertically upwards from the surface of a spherical planet of radius $R$ and mass $M$ with an initial speed equal to half of the planet's escape velocity. Neglecting atmospheric friction, what maximum height above the planet's surface will the object reach?

\frac{R}{4}

\frac{R}{3}

\frac{R}{2}

R

Show answer & explanation

Answer: $\frac{R}{3}$

Answer

The maximum height above the planet's surface reached by the object is

\frac{R}{3}

.

By mechanical energy conservation, the initial total energy at launch equals the potential energy at maximum height where velocity is zero. The launch velocity

v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}

gives an initial kinetic energy of

\frac{GMm}{4R}

. Combined with initial potential energy

-\frac{GMm}{R}

, total energy is

-\frac{3GMm}{4R}

. Equating this to

-\frac{GMm}{R+h}

gives

R+h = \frac{4}{3}R

, yielding a height above the surface of

h = \frac{R}{3}

.

Step-by-Step Solution

1

Express the initial speed in terms of gravitational constant

G

, mass

M

, and radius

R

.

The escape velocity is

v_e = \sqrt{\frac{2GM}{R}}

. Thus, launch speed

v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}

.

Escape velocity is defined as the minimum speed needed to escape the gravitational field.

2

Calculate the initial total mechanical energy at the planet's surface.

Initial kinetic energy

K_i = \frac{1}{2}m v^2 = \frac{1}{2}m\left(\frac{2GM}{4R}\right) = \frac{GMm}{4R}

. Surface potential energy

U_i = -\frac{GMm}{R}

. Total energy

E_i = K_i + U_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}

.

Total mechanical energy is the sum of kinetic energy and gravitational potential energy.

3

Apply energy conservation to find the maximum distance from the planet's center.

At maximum height

h

, speed is zero (

K_f = 0

), so distance from center is

r = R + h

. Energy

E_f = -\frac{GMm}{R+h}

. Equating

E_i = E_f

gives

-\frac{3GMm}{4R} = -\frac{GMm}{R+h} \implies R+h = \frac{4}{3}R \implies h = \frac{R}{3}

.

Mechanical energy is conserved in a central gravitational force field.

Key Concept

Conservation of Mechanical Energy in a Gravitational Field

Estimated Time:1m 30s

Question 13Question →

A satellite moves in a circular orbit of radius $2R$ around a spherical planet of radius $R$ and mass $M$ . It is subsequently shifted to a larger circular orbit of radius $8R$ . What is the ratio of its initial orbital speed to its new orbital speed?

4 : 1

2 : 1

1 : 2

1 : 4

Show answer & explanation

Answer: $2 : 1$

Answer

The ratio of the initial orbital speed to the new orbital speed is

2 : 1

.

Orbital velocity varies inversely with the square root of orbital radius (

v \propto \frac{1}{\sqrt{r}}

). Moving from radius

2R

to

8R

decreases the speed by a factor of

\sqrt{8/2} = 2

. Therefore, the ratio of initial speed to new speed is

2 : 1

.

Step-by-Step Solution

1

Write the formula for orbital velocity

v = \sqrt{\frac{GM}{r}}

, where

G

is the gravitational constant,

M

is the mass of the planet, and

r

is the orbital radius.

Orbital speed is determined by equating gravitational force to centripetal force.

2

Set up expressions for initial and final orbital speeds

Initial speed

v_1 = \sqrt{\frac{GM}{2R}}

and final speed

v_2 = \sqrt{\frac{GM}{8R}}

.

Substitute the given orbital radii

r_1 = 2R

and

r_2 = 8R

into the formula.

3

Calculate the ratio

v_1 / v_2

\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{GM}{8R}}} = \sqrt{\frac{8R}{2R}} = \sqrt{4} = 2

Simplifying the square root fraction gives the exact ratio.

Key Concept

Orbital Velocity and Inverse Square Law Relations

Estimated Time:1m 15s

Question 14Question →

Two point masses, each of mass $125\text{ kg}$ , are fixed at positions $(-3.0\text{ m}, 0)$ and $(3.0\text{ m}, 0)$ in the $xy$ -plane. What is the magnitude of the net gravitational field strength at a point $P$ located at $(0, 4.0\text{ m})$ , in terms of the universal gravitational constant $G$ ?

8.0G\text{ N/kg}

10.0G\text{ N/kg}

6.0G\text{ N/kg}

4.0G\text{ N/kg}

Show answer & explanation

Answer: $8.0G\text{ N/kg}$

Answer

The magnitude of the net gravitational field strength at point

P

is

8.0G\text{ N/kg}

.

The distance from each mass to the target point is 5.0 m, yielding an individual field magnitude of 5.0G N/kg. Resolving vectors shows that horizontal components cancel out while vertical components add to give 2 × (5.0G × 4/5) = 8.0G N/kg.

Step-by-Step Solution

1

Calculate the distance

r

from each mass to point

P(0, 4.0\text{ m})

.

r = \sqrt{(-3.0 - 0)^2 + (0 - 4.0)^2} = \sqrt{9 + 16} = 5.0\text{ m}

.

The distance formula in two dimensions gives the hypotenuse of the right triangle formed by the coordinates.

2

Determine the magnitude of the gravitational field

E

created by each individual mass at point

P

.

E = \frac{G M}{r^2} = \frac{G \times 125}{5.0^2} = \frac{125G}{25} = 5.0G\text{ N/kg}

.

Newton's law of universal gravitation defines field strength as

E = \frac{GM}{r^2}

.

3

Resolve the field vectors into horizontal and vertical components.

By symmetry, the horizontal components

E_x = E \sin\theta

are equal in magnitude and opposite in direction, so

E_{x,\text{net}} = 0

. The vertical components

E_y = E \cos\theta

point downwards towards the origin.

Gravitational field is a vector quantity, so opposite components cancel while aligned components add together.

4

Calculate the total vertical component of the net gravitational field.

\cos\theta = \frac{4.0}{5.0} = 0.8

. Thus,

E_{\text{net}} = 2 \times E_y = 2 \times (5.0G \times 0.8) = 8.0G\text{ N/kg}

.

Summing the vertical contributions from both identical masses gives the net magnitude.

Key Concept

Vector Addition of Gravitational Field Strengths

Question 15Question →

A planet has a mass equal to $8$ times the mass of the Earth and a radius equal to $2$ times the radius of the Earth. If the acceleration due to gravity on the surface of the Earth is $10\text{ m/s}^2$ , what is the acceleration due to gravity on the surface of the planet in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 20

Answer

The acceleration due to gravity on the surface of the planet is

20\text{ m/s}^2

.

The acceleration due to gravity at the surface of a spherical body is given by

g = \frac{GM}{R^2}

. When the mass is multiplied by

8

and the radius is multiplied by

2

, the new acceleration becomes

g_p = \frac{G(8M)}{(2R)^2} = \frac{8}{4}\frac{GM}{R^2} = 2g

. Substituting

g = 10\text{ m/s}^2

gives

20\text{ m/s}^2

.

Step-by-Step Solution

1

Write the expression for acceleration due to gravity at the surface of Earth.

g_e = \frac{GM_e}{R_e^2} = 10\text{ m/s}^2

Gravitational field strength at the surface depends directly on body mass and inversely on the square of radius.

2

Set up the ratio equation for the planet's gravitational acceleration using relative mass and radius values.

g_p = \frac{G(8M_e)}{(2R_e)^2} = \frac{8}{4} \left(\frac{GM_e}{R_e^2}\right) = 2g_e

Squaring the radius multiplier of

2

yields

4

in the denominator, while the numerator increases by a factor of

8

.

3

Calculate the final numeric value.

g_p = 2 \times 10\text{ m/s}^2 = 20\text{ m/s}^2

Multiplying Earth's value by the combined ratio of

2

gives the answer.

Key Concept

Gravitational field strength on planetary surfaces (

g = \frac{GM}{R^2}

)

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